text:
\section{A multisingular hyperbolic set in $M^5$}\label{s.main}
The aim of this section is to find a chain recurrent set that is multisingular hyperbolic with 2 singularities of different indexes.
 For this, the strategy will be to multiply the vector field $X$ in $S^3$ from section  \ref{attractor-repeller} times  $\mathbb{RP}^2$ with a simple vector field. Then modify the resulting set to obtain new recurrence.

The following Lemma will be proven in subsection\ref{ss.propY}.
\begin{lemm}\label{propY}
There exist a vector field $Y$ in $\mathbb{RP}^2$ with the following properties:
\begin{itemize}
\item $Y$ is a $\mathcal{C}^{\infty}$ vector field

\item It has 3 singularities: a saddle singularity $s$, a source $\alpha$ and a sink $\omega$.
\item The contracting and expanding  Lyapunov exponents of the saddle are equal in absolute value ($\lambda_{sss}=-\lambda_{uuu}$), and $\lambda_{uuu}>> 6$.
\item One of the stable branches of $s$ (that is an orbit) has $\alpha$-limit  $\alpha$.
and one of the unstable branches of $s$ (that is an orbit) has $\omega$-limit   $\omega$.
\item The other two branches form an  orbit with $\alpha$-limit and $\omega$-limit in $s$ and we call this orbit $\gamma$.
\item There is a transverse section to $\gamma$ and to the flow, that we call $T=[-1,1]\times\set{a}$,  $T\cap\gamma=0\times{a}$ and the flow of $Y$, $\phi^Y(s,t)$ is such that:
    \begin{itemize}
    \item  If $s=(x,a)\in T$ is such that $x>0$, then $\phi^Y(s,t)$  does not cross $T$  for any $t>0$ and has $\omega$-limit in $\omega$. And for $t<0$ there exists only one $t_s<0$ such that $\phi^Y(s,t_s)=s'\in T$ with $s'=(x',a)$, $x'<0$ and the $\alpha$-limit of $s$ is $\alpha$.
    \item If $s=(x,a)\in T$  and $x<0$, then  $\phi^Y(s,t)$ does not cross $T$  for any $t<0$ and has $\alpha$-limit in $\alpha$. And for $t>0$ there exists only one $t_s>0$ such that $\phi^Y(s,t_s)=s'\in T$ with $s'=(x',a)$, $x'>0$ and the $\omega$-limit of $s$ is $\omega$.

    \end{itemize}
\end{itemize}
\end{lemm}

\subsection{The vector field in $M^5$}
We start by considering the vector field $Z_{id}=(X,Y)$ in the manifold $M^5=S^3\times\mathbb{RP}^2$ and it's flow $\phi_{id}$ .
Let us define the section $$\sum=S^3\times T$$ which is transverse to $Z_{id}$, and a flow-box $\sum\times[-1,0]$.
\begin{prop}
Let $H:\sum\to\sum$ be a $\mathcal{C}^{\infty}$ diffeomorphism isotopic to identity and that is the identity on the boundary.
There exist a  flow $\phi_{H} $ in  $M^5$, associated to a $\mathcal{C}^{1}$ vector field $Z_H$, such that $\phi_H=\phi_{id}$ in the complement of the flow-box $\sum\times[-1,0]$, and in the flow-box $(H(z),0)=\phi_H((z,-1),1)$.

\end{prop}
\proof
Since $H $ is isotopic to the identity we have that there exist a diffeomorphism $F:\sum\times[-1,0]\to\sum$ such that $F(\sum, -1)=id$ and $F(\sum, 0)=H$.
We also have that there exist $F':\sum\times[-1,0]\to\sum$ such that $F'(\sum, -1)=H^{-1}$ and $F'(\sum, 0)=id$.
Let us define the flow $\phi_H$ as follows:

\begin{itemize}
\item $\phi_H(y,t)=\phi_{id}(y,t) $ for every $t$ such that $\phi_H(y,t)\notin \sum\times[-1,0]$
\item If $t_0$ is such that $\phi_H(y,t_0)\in  \sum\times\set{-1}$ then $$\phi_H(y,t)=F(\phi_{id}(y,t_0),s)\,,$$  for every  $s=t-1-t_0$  such that $-1\leq  s \leq0$.
\item If $t_1$ is such that $\phi_H(y,t_1)\in  \sum\times\set{0}$ then $$\phi_H(y,t)=F'(\phi_{id}(y,t_1),s)\,,$$  for every  $-s=t-t_1$  such that $-1\leq  s \leq0$.

\end{itemize}
Now we define the vector field $Z_H$ by taking at any point,  the derivative (on $t$) of $\phi_H(y,t)$ and since $\phi_H(y,t)$ is sufficiently smooth, then so  is $Z_H$.
\endproof

\subsubsection{ A filtrating region for $Z_H$ }

We recall that $U$ is the filtrating region in $S^3$ for $X$, defined in section \ref{attractor-repeller}.

We define now the filtrating region in $M^5$ that is interesting to us:
We consider a repelling region $u_{\alpha}\subset \mathbb{RP}^2$ of $\alpha$ for $Y$, such that $\alpha$ is the maximal invariant set in $u_\alpha$. Similarly, consider  a trapping region $u_{\omega}\subset \mathbb{RP}^2$
We take  the respective repelling and trapping regions of this singularities in $M^5$. We define the repelling region $U_{\alpha}=S^3\times u_{\alpha}$ and the trapping region $U_{\omega}=S^3\times u_{\omega}$.
We define as well $$V=[U\times \mathbb{RP}^2]/\set{U_{\alpha}\cup U_{\omega}}\,.$$

Let us consider the maximal invariant set for $Z_{id}$ in $V$ that we call $\Lambda_{id}$.
\begin{prop}\label{p.max1}
The maximal invariant set $\Lambda_{id}$ in $V$ (for $Z_{id}$) intersects $\sum$. For any $H $ as above, any orbit in the maximal invariant set $\Lambda_H\in V$ (for $Z_{H}$) either crosses $\sum$ or is contained in  $S^3\times \set{s}$.
\end{prop}
\proof
Let us consider the saddle singularity in $Y$ that we called  $s$. By construction, there is a unique orbit of $Y$, formed by  a branch of the stable and unstable manifold of $s$,  that crosses $T$. Since the contraction and expansion rates in $Y$ are stronger than in $X$, then the points in $S^3\times\set{s}$ have a connection between the strong stable and unstable manifolds and the orbits in this connections cross $\sum$.

If the orbit $\gamma_y $ of a point $y=\set{(x,l)}$ never crosses $\sum$ then $$Z_{id}\mid_{\gamma_y}=Z_{H}\mid_{\gamma_y}.$$
Let us see that $\Lambda_{id}$ is contained in  $S^3\times \set{s}$ or it crosses $\sum$.

We take $u_0=\mathbb{RP}^2/u_{\alpha}\cup u_{\omega}$. Then the maximal invariant set in $u_0$ for $Y$ is the saddle $s$ and the saddle connection (the orbit that contains one unstable branch and one stable branch of $s$). All other points have their $\alpha$ and $\omega$-limits in the singularities $\alpha$ and  $\omega$ (see the properties of $Y $ in (\ref{propY})).
So if there is a point $y\in\gamma_y$ such that $y\notin S^3\times \set{s} $ and $\gamma_y\cap \sum=\emptyset$ then the orbit of $l$ by $Y$ has  $\alpha$ or $\omega$-limits in the singularities $\alpha$ and  $\omega$. This implies that $y$ has $\alpha$ and $\omega$-limits in $U_{\alpha}\cup U_{\omega}$. Therefore $\gamma_y\notin \Lambda_{id}$.
\endproof

 \subsection{Some usefull subsets of $S^3$}

Recall that there are 2 saddles singularities in $S^3$, $\sigma_a\subset L_a$ and $\sigma_r \subset L_r$.
Also from the construction of $X$, the wondering orbits with $\omega$-limit in $L_a$ had three possible kinds of $\alpha$-limit
\begin{itemize}
\item a  source called $p_1'$ or $p_1$
\item a chain recurrence class that is just a periodic orbit
\item a subset of $L_r$.
\end{itemize}
Analogously,  the wondering orbits with $\alpha$-limit in $L_r$ had three possible kinds of $\omega$-limit
\begin{itemize}
\item a sink called $p_2'$ or $p_2$
\item a chain recurrence class that is just a periodic orbit
\item a subset of $L_a$.
\end{itemize}

\begin{defi}

\begin{itemize}

\item We define $K_Y+ 1$ as the minimum of the times  for a point in $T$ to return to $T$ for $Y$.
\item By construction of the Lorenz attractor (see \cite{GuWi}) there is a small linear neighborhood around the singularity $\sigma_a$, in which we can consider the coordinates $(x,y,z)$ to correspond to the strong unstable, weak stable and stable spaces respectively. We can choose thees coordinates so that singularity is approached by orbits of $L_a$ only in one semi space that corresponds to the points with positive $y$ value. We say then that $\sigma_a$ has an \emph{escaping separatrix} $W^{cs-}$ which is the half stable manifold that escapes from a neighborhood of $L_a$.

  In the same way there is an escaping separatrix $W^{cu+}$ for the singularity $\sigma_r$ in $L_r$.
\item We  choose an orbit $\gamma_{cs} \in W^{cs-}$  such that $\gamma_{cs}$ is tangent to $y$ in the linearized neighborhood of $\sigma_a$ and is in the basin of attraction of a source (we choos $p_1'$ for convenience) for the past. There is a neighborhood of   $\gamma_{cs}$ (with $\sigma_a$ in the boundary)  that we call $D_a$  that is a repelling region.
In the same way we define $D_r$. By choosing these neighborhoods smaller  $D_a\cup D_r$ does not disconnect  $S^3$.
\item We define  the corresponding repelling region in $M^5$, $V_r=\mathbb{RP}^2\times D_r.$
\item We define a compact ball $B_a\subset S^3 $ that intersects $ U$  in a linearized region  of the singularity $\sigma_a$ and it's boundary  intersects $L_a$ only in one point $p$ of the local unstable manifold of $\sigma_a$.
In other words, inthe coordinates given above if
$ (x,y,z)\in B_a$ then $y\geq0$ and it is $0$ only in $p\in W_X^u(\sigma_a)$.
\item  We define  a compact ball $B_r\subset U$ in the linearized neighborhood of $\sigma_r$,such that

if $ (x,y,z)\in B_r$ then $y\leq0$ and it is $0$ only in $p'\in W_X^s(\sigma_r)$.
The ball $B_r$  and $D_r$ can be taken so that there exist $t_0<K_Y$ such that $\phi_X^{t}(B_r)\subset D_r$  for all $t>t_0$.
 Note that here the stable and unstable manifolds refer to de dynamics of $X$.

\end{itemize}
\end{defi}
\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.60\linewidth]{bolasinper1.pdf}
\end{center}
\caption{The ball $D_a$.}
\end{figure}

 \subsection{Choosing the right isotopy}\label{s.isotopy}
  We  now
choose some more properties on the diffeomorphism $H$ from proposition \ref{p.max1}.
The following Lemma will be proven in subsection \ref{s.conH}.
\begin{lemm}\label{l.propH}
Let $\Sigma\subset S^3 \times \mathbb{RP}^2$ be $\sum=S^3\times T$ where $T\subset  \mathbb{RP}^2$ is a uni-dimensional  transverse section to the vector field $Y$ parametrized by  $[-1,1]$.
There exist a $\mathcal{C}^{\infty}$ diffeomorphism isotopic to identity,  $H:\Sigma\to\Sigma$,
$H(x,l)=(r_l(x),\theta_x(l))$
 where $r_l: \sum\to S^3$, $\theta_x:\sum\to T $, with  following properties:
\begin{itemize}

\item $H$ is the identity on the boundary of $\sum$  and  outside of $V_a$ and in $V_r$.
\item The map $r_l(x)$ is the identity for $l=1$ or  $l=-1$, or if $x\in \overline{D_r}$.
\item
The image of $r_l(B_a)=B_r$, and $r_l(p)=p'$ for all $l\in[-1/2,1/2]$.
\item  If $l\in[-1/2,1/2]^c$ then $\theta_x(l)=l$.
\item  If $l\in[0,1/2]$ and $x\notin B_a$ $\theta_x(l)>0$ .
\item  If $l\in[0,1/2]$ and $x\in B_a$ then $-\epsilon\leq\theta_x(l)\leq \epsilon$,

\item  The only point $l$ such that $H(p,l)=(p',0)$, is $l=0$.
\item $H(S^3\times\set{0})\times\set{0}\in\Sigma\times\set{0}$$  cuts transversally $$S^3\times\set{0}\times\set{0}\in\Sigma\times\set{0}\,.$
\end{itemize}
\end{lemm}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.65\linewidth]{paste1.pdf}
\end{center}
\caption{The function $\theta_x(l)$. Note that the painted area depicts orbits with their $\alpha$ limits in $U_{\alpha}$ and the striped area depicts orbits with their $\omega$ limits in $U_{\omega}$}
\end{figure}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.70\linewidth]{bolasinper.pdf}
\end{center}
\caption{The ball $B_a$}
\end{figure}

\begin{prop}\label{p.max2}
We consider $H$ as in Lemma \ref{l.propH}, then the orbits in the maximal invariant set $\Lambda_H$ are contained in $L_a\times \set{s}$ or cross the flow box $\sum\times [-1,0]$ in $$B_a\times [0,1/2]\times\set{-1}\,.$$

\end{prop}
\proof
Suppose that $\gamma $ is an orbit in $\Lambda_H$ that doesn't cross $\sum$. From Proposition $\ref{p.max1}$ these orbits of $\Lambda_{H}$ are in $S^3\times\set{s}$.
Let $h $  be a point of $\gamma$ of coordinates $(x,j)\in S^3\times\mathbb{RP}^2$. If $x$ is not in $\La$ (for $X$) then, the alpha or the omega limit of $x$ must be in $U^c$. Therefore, for a $t$ large enough, $\phi^t_{H}(x,j)\notin V$ ( Recall that $V=[U\times \mathbb{RP}^2]/\set{U_{\alpha}\cup U_{\omega}}\,$).

Then if $\gamma$  doesn't intersect $\sum$, it must be in $\La\times \set{s}$.

Let us suppose now that $\gamma $ intersects $\sum$. Let $h$ be a point in $\gamma\cap\sum\times[-1,0]$ such that $h\in \sum\times\set{-1}$. We write $h$ as $(y,-1)$ and $y$ as $y=(x,l)\in\sum$.
\begin{enumerate}
\item If $l>1/2$, or if $x\notin B_a$ with $l>0$, then $H(x,l)=(r_l(x),\theta(l))$ with $\theta(l)>0$ and then $\phi^{1}_H(y)=(r_l(x),\theta(l))\times\set{0}$. Since outside of the flow-box $Z_{id}=Z_{H}$ now we can look at $Z_{id}$. From the properties of  $Y$ (\ref{propY}) we have that the future orbit of  $\theta(l)>0$, does not cross $T$  and the $\omega$-limit   is $\omega$.
    Then the orbit for $\phi^{t}_H$ is in $U_{\omega}$ for a large enough $t$.
    Then $\gamma$ is not in $\Lambda_H $.
\item  If $l<0$, since $h$ goes  outside of the flow-box for the past (where $Z_{id}=Z_{H}$) now we can look at $Z_{id}$. From the properties of  $Y$ (\ref{propY}) we have that the orbit of  $l<0$ does not cross $T$ for the past and the $\alpha$-limit  is $\alpha$. Then $\gamma$ does not cross again the flow-box for the past. The orbit for $\phi^{t}_H(h)$ is in $U_{\alpha}$ for a negatively large enough $t$ and $\gamma$ is not in $\Lambda_H $.

\item If $x$ is not in  $B_a$ and $l=0$ then $\phi^{1}_H(h)=((H(x),\theta_x(l)),0)$ and $\theta_x(l)>0$. Then, as before,  we have that the orbit of $\theta_x(l)$ for $Y$ does not cross $T$ for the future and the $\omega$-limit  for $Y$ is $\omega$. Then $\gamma$ does not cross again the flow-box for the future and $\gamma$ is not in $\Lambda_H $.

 \end{enumerate}
Then the only other case in which $\gamma$ might be in $\Lambda_H $ is if $\gamma$ crosses  the flow box $\sum\times[-1,0]$ in $B_a\times [0,1/2]\times\set{-1}$.
 \endproof

\begin{prop}\label{p.max3}
There is a unique orbit $\gamma$ in $\Lambda_H$ that crosses $\sum$, that orbit is the orbit of $$(p,0)\times \set{-1}\in \sum\times\set{-1}\,.$$
\end{prop}
\proof
Let $\gamma$ be an orbit in $\Lambda_H$.
From proposition (\ref{p.max2}), we already know that if an orbit of $\Lambda_H$ crosses $\sum$ then it crosses at a point $y=(x,l,-1)\in B_a\times [0,1/2]\times\set{-1}$.

If $\theta_x(l)<-\epsilon<0$ recall that the properties of $H$ (Lemma \ref{l.propH}) give us that  then $l<0$.

Suppose now that $l\geq0$ and that $\theta_x(l)>0$. As in our previous proposition this implies that $\phi_H^{t+1}(y)\notin U$ for $t$ large enough.

If $l\geq0$ and $-\epsilon\leq\theta_x(l)\leq 0$  then $x\in B_a$, and  $r_l(x)\in B_r$. Suppose that $x\neq p$.
 Then $$\phi_H^{t+1}(y)\notin \sum\times[-1,0]$$ for all $K_Y>t>0$, and therefore $\phi_H^{t+1}(y)=\phi_{id}^{t}(\phi_H^{1}(y))$ for all $K_Y>t>0$.
 Let us consider $t_0$ as in the properties of $H$ (Lemma \ref{l.propH}). Recall that $t_0 $ is such that  $\phi_X^{t}(B_r)\subset\overline{D_r}$  for all $K_Y>t>t_0$.
We  call  $$\phi_{id}^{t_0}(\phi_H^{1}(y))=(x_1,z_1)\in S^3\times\mathbb{RP}^2\,.$$
 Since $x_1\in u_r$ and is not in $L_r$ (since $x_1\neq p'$), then   $x_1$ is in the  attracting region of a sink  $p_r$, $D_r$  of $X$ (see subsection \ref{completeflow3}).
 Now, for all $t>t_0$ even the ones bigger than $K_Y$,  we have that  $$\phi_H^{t}(x_1,z_1)=(\phi_{Hx}^{t}(x_1,z_1),\phi_{Hz}^{t}(x_1,z_1))$$ and since every time for the future that this orbit crosses the flow-box $\sum\times[-1,0]$, the function $r_l$ is the identity (since  $r_l$ is the identity in $D_r$ ), then $$\phi_H^{t}(x_1,z_1)=(\phi_{X}^{t}(x_1),\phi_{Hz}^{}(x_1,z_1))\,.$$

 Since
 $\phi_X^{t}(x_1)\notin U$ for $t>t_0$ big enough,  then  $\phi_H^{t'}(y) $ is eventually not in $V$ for some   $t>t_0$. Then  $\gamma$ is not in $\Lambda_H$ as wanted.

If $l\geq0$ and $-\epsilon\leq\theta_x(l)< 0$ but $x=p$.
Let $t_y$ be a time in which the orbit returns to the flow-box. That is $t_y$ is such that $$\phi_H^{t_y}(y)=(x_1,l_1,-1)\in \sum \times\set{-1}\,.$$
  Recall from  the properties of $H$ (Lemma \ref{l.propH}) that $t_y\geq K_Y >t_0$ with $t_0$  such that  $\phi_X^{t}(B_r)\subset \overline{D_r}$  for all $t>t_0$.
 Since$-\epsilon\leq\theta_x(l)<0$ and after returning to $\sum \times\set{-1}\,$ the orientation was reversed, then $l_1$ is positive. Since now $x_1$ is not in $B_a$, then $\theta_{x_1}(l_1)>0$. So, now for a small $t>0$, we have that  $$\phi_H^{t+t_y+1}(y)=\phi_{id}^{t}(\phi_H^{t_y+1}(y))\,.$$ This implies that the orbit of $y $ never cuts the flow box again, and therefore, for a big enough $t$, $\phi_H^{t}(y)$ is in $U_{\omega}$. As a consequence  $\gamma$ is not in $\Lambda_H$ as wanted.

The only case left is  $x=p$ and $\theta_x(l)=0$. The last property of $H$ (Lemma \ref{l.propH}) tells us that  $l=0$, so the objective now is to prove that the orbit of
$y=(p,0) $ never leaves $$V=[U\times \mathbb{RP}^2]/\set{U_{\alpha}\cup U_{\omega}}\,.$$
But  $(p,0) $ is in the stable manifold of $\sigma_r$ and in the unstable manifold of $\sigma_a$, and then then the orbit of $y$ is in $\La_H$..

\endproof

\subsection{Multisingular hyperbolicity}
Until now we have constructed a vector field having a chain recurrent class such that  \begin{itemize}
                                                       \item Two singularities of different indexes one in $L_a$ and the other in $L_r$.
                                                       \item All the periodic orbits have the same index and the singularities are in the closure of the periodic orbits.
                                                       \item There are  periodic orbits in $L_a$ such that their  stable manifolds intersect  the unstable manifolds of periodic orbits in $L_r$.
                                                       \item There is only one orbit in the class with the $\alpha$-limit in $L_a$ and the $\omega$-limit in $L_r$.
                                                     \end{itemize}
The goal now is to show that we can choose a diffeomorphism $H $ so that this vector field would be strong multisingular hyperbolic.
After that we will perturb this vector field to an other that will still be strong multisingular hyperbolic, but having a homoclinic connection between periodic orbits in $L_a$ and periodic orbits in $L_r$. This will finish the proof of theorem \ref{t.example}.

In subsection \ref{s.conH} we will prove that there exist a diffeomorphism $H $ with the properties defined in  Lemma \ref{l.propH}, in particular   with the following  property, we require that $$H(S^3\times\set{0})\times\set{0}\in\Sigma\times\set{0}$$  cuts transversally $$S^3\times\set{0}\times\set{0}\in\Sigma\times\set{0}\,.$$

This last property guaranties that the set  $\Lambda_H$  will be strong multisingular hyperbolic.

To show that $\Lambda_H$  is strong multisingular hyperbolic we need to check that we are in the hypothesis of Lemma \ref{c.unamashyp}. Since we have already shown the other hypothesis
the following Lemma implies strong multisingular hyperbolicity.
\begin{lemm}\label{l.unaorbitamas}
Let $y\in\Sigma\times\set{-1}$ be such that $\Lambda_H=\Lambda\cup O(y)$.
There exist a diffeomorphism $H$ such that \begin{itemize}
                                             \item The stable and unstable spaces along the orbit of $S_{Z_H}(y)$ intersect transversally,
                                             \item The orbit of $y$ does not intersect the escaping spaces of the singularities for $Z_H$,
                                           \end{itemize}
Moreover this implies that  $\Lambda_H$ is multisingular hyperbolic.

\end{lemm}

\proof
Consider the points in $S^3\times \mathbb{RP}^2$, $a=(p,s)$ and $b=(p',s)$ and  $y=(p,0,-1)\in\Sigma\times \set{-1}$.
The orbit of $y$ is in the strong unstable manifold of $a$, (since  the unstable manifold of $s$ intersects $T$ at $0$ for $Y$ ). Analogously $y$ is in the strong stable manifold of $b$ since $\phi_H^{1}(y)=(p',0,0)$. Observe that $a$ and $b$ are regular points and $p\in W^u(\sigma_a)$ and $p'\in W^s(\sigma_r)$ for $X$. therefore $\gamma$ does not intersect the escaping spaces of the singularities for $Z_H$. From Proposition \ref{p.unamasextended} this implies that the center space of the singularities of  $\Lambda_H$ and $\Lambda$ are the same.

From Lemma \ref{l.masuna} we have that there exists an unstable space (for the reparametrized linear Poincar\'{e} flow ) at $y$ that we call $E^u_y$. We can consider a metric so that $\Sigma$ is always  normal
to the vector field at the flow box.
We take a vector $v\in E^u_y$ at $y$. This vector is tangent to  $$S^3\times\set{0}\times\set{-1}\in\Sigma\times\set{-1}$$ at $y$.

Let us recall that we have assumed at the beginning of the subsection that the image of $$H(S^3\times\set{0})\times\set{0}\in\Sigma\times\set{0}$$ under $H$ cuts transversely $$S^3\times\set{0}\times\set{0}\in\Sigma\times\set{0}\,.$$ Then the image of $v$ under the differential of $H$ (and of $\phi_H^{1}$) is transverse to $S^3\times 0\times O(y)$ at $\phi_H^{1}(y)$, and then so is the image of $v$ under $\Psi^1(v)$, since the direction of the flow is perpendicular to $T\times\set{0}$.
On the other hand Lemma \ref{l.masuna} also gives us a  stable space $E^s_y$ at  $\phi_H^{1}(y)$  that is tangent to  $S^3\times \set{0}\times\set{0} $ at $\phi_H^{1}(y)$. Then the stable and unstable spaces of the reparametrized linear Poincar\'{e} flow are transverse. Then  we are in the hypothesis of Corolary \ref{c.unamashyp} and this completes the proof.
  \endproof

 With this last Lemma we know that the maximal invariant set $\La_H$ is multisingular hyperbolic. But this is not enough, since a small perturbation of $Z_H$ could brake the connection between $L_a$ and $L_r$ and have $\sigma_a$ and $\sigma_r$ in different chain classes.
 We need now to show that the right perturbation of   $Z_H$  will generate  intersection of the stable and unstable manifolds of periodic orbits in $L_a$ and $L_r$. Since $\La_H$ is multisingular hyperbolic for $Z_H$, so will it be for this new vector field and now the singularities will be robustly in the same chain recurrence class.

  The following Lemma implies Theorem \ref{t.example}
  \begin{figure}[htb]
\begin{center}
\includegraphics[width=0.70\linewidth]{bolasper.pdf}
\end{center}
\caption{A perturbation of $Z_H$, in particular of $r_l(x)$.}
\end{figure}
\begin{lemm}\label{perint}
  There is an arbitrarily small perturbation of $Z_H$, that we call $Z_{H_{\epsilon}}$, and a $C^1$ neighborhood of $Z_{H_{\epsilon}}$ called $\cV$ so that any vector field $Z\in\cV$ has a maximal invariant set $\La_Z$ that is multisingular hyperbolic and there is a chain class $C\subset\La_Z$ that has two singularities of different index accumulated by periodic orbits.

\end{lemm}

\proof
We will make a small perturbation of $H$ and this will result in a small perturbation of $Z_H$.
Let us recall that we can write $H$ as
$$H(x,l)=(r_l(x),\theta_x(l))$$ where $r_l: \sum\to S^3$, $\theta_x:\sum\to T $.

Let us again choose coordinates in the linearized neighborhood of $\sigma_r$ as the chosen before. Consider a small positive traslation along the $y$ direction of the ball $B_r$, $T_\epsilon(B_r)$ so that there exists  $b_r\subset{B_r}$  a small ball such that the periodic orbits of $L_r $ are dense in $b_r$ and has $T_\epsilon (p')$ in it's boundary. Analogously for $B_a$
We can perturb $r_l$ to $r'_l(x)$ so that $T_\epsilon(B_a)$ is sent to $T_\epsilon(B_r)$ and  $r'_l(T_\epsilon(p))$ is now sent to $T_\epsilon(p')$.
By choosing $\epsilon$ small enough this can be achived with a $C^r$ small perturbation of $r_l$.
The resulting vector field  $Z_{H_{\epsilon}}$ is still $C^1$ and multisingular hyperbolic.

Now from the fact that periodic orbits are dense in the sets $L_a$ and $L_r$, and the fact that  $Z_{H_{\epsilon}}$ is star, we get that we can choose a small perturbation by the connecting Lemma in [\cite{BC}] so that the  some periodic  orbit in $L_a$ is homoclinically related to a periodic orbit in $L_r$. Recall that the periodic orbits all have the same index.
This homoclinic intersection is roust.
\endproof

\section{Appendix: Sub constructions}

\subsection{ Construction of the vector field $Y$ in $\mathbb{RP}^2$ }\label{ss.propY}

In this subsection we prove the following Lemma from  section (\ref{s.main})
\begin{lemm*}
There exist a vector field $Y$ in $\mathbb{RP}^2$ with the following properties:
\begin{itemize}
\item $Y$ is a $\mathcal{C}^{\infty}$ vector field

\item It has 3 singularities: a saddle singularity $s$, a source $\alpha$ and a sink $\omega$.
\item The contracting and expanding  Lyapunov exponents of the saddle are equal in absolute value ($\lambda_{sss}=-\lambda_{uuu}$), and $\lambda_{uuu}>> 6$.
\item One of the stable branches of $s$ (that is an orbit) has $\alpha$-limit  $\alpha$,
and one of the unstable branches of $s$ (that is an orbit) has $\omega$-limit   $\omega$.
\item The other two branches form an  orbit with $\alpha$-limit and $\omega$-limit in $s$ and we call this orbit $\gamma$.
\item There is a transverse section to $\gamma$ and to the flow, that we call $T=[-1,1]\times\set{a}$,  $T\cap\gamma=0\times{a}$ and the flow of $Y$, $\phi^Y(s,t)$ is such that:
    \begin{itemize}
    \item  If $s=(x,a)\in T$ is such that $x>0$, then $\phi^Y(s,t)$  does not cross $T$  for any $t>0$ and has $\omega$-limit in $\omega$. And for $t<0$ there exists only one $t_s<0$ such that $\phi^Y(s,t_s)=s'\in T$ with $s'=(x',a)$, $x'<0$ and the $\alpha$-limit of $s$ is $\alpha$.
    \item If $s=(x,a)\in T$  and $x<0$, then  $\phi^Y(s,t)$ does not cross $T$  for any $t<0$ and has $\alpha$-limit in $\alpha$. And for $t>0$ there exists only one $t_s>0$ such that $\phi^Y(s,t_s)=s'\in T$ with $s'=(x',a)$, $x'>0$ and the $\omega$-limit of $s$ is $\omega$.

    \end{itemize}
\end{itemize}
\end{lemm*}
 \subsubsection{ A vector field with a saddle connection in a M\"{o}bius strip }

 Let us start by defining some simple linear flow in $\RR^2$.
 We take a linear vector field $Y(x,y)=(\lambda_{sss}x, \lambda_{uuu}y )$ defined in $[-2,2]\times[-2,2]$. We ask that $\lambda_{uuu}=-\lambda_{sss}$ and we also ask that  $\lambda_{uuu}>6$.

We consider a close curve $C$ formed by the union of following curves:
\begin{itemize}
\item We consider the orbit of  a point $(-a,2)$. This orbit cuts the vertical line $(-2,y)$ in a point $(-2,a')$. The segment of orbit from $(-2,a')$ to $(-a,2)$ is our first curve $C_1$.
\item  We consider the orbit of  a point $(a,2)$. This orbit cuts the vertical line  $(2,y)$ in a point $(2,c)$. The segment of orbit from $(a,2)$ to $(2,c)$ is  $C_2$.
\item We consider the segment $\set{-2}\times[a',-a']$ as our second curve  $C_3$.

\item We take the  orbit of $(-2,-a')$ and we call the point where it cuts the horizontal line $l$ in a point $(-b,-2)$. The segment of orbit from $(-2,-a')$ to $(-b,-2)$ is our third curve  $C_4$.
\item We consider the segment $\set{2}\times[-c,c]$ as our second curve  $C_5$.

\item  We consider the orbit of  a point $(2,-c)$. This orbit cuts the horizontal line  $(x,-2)$ in a point $(b',-2)$. The segment of orbit from $(2,-c)$ to $(b',-2)$ is  $C_6$.
\item The segment  $[b',-b]\times\set{-2}$ our forth curve  $C_7$.

  \item The segment  $[-a,a]\times\set{2}$ our last curve  $C_8$.
\end{itemize}
\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.70\linewidth]{vectroY.pdf}
\end{center}
\caption{The vector field $Y$ in $D$.}
\end{figure}

There is a diffeomorphism that reverse orientation $d:C_8\to C_5$ defined as follows: $$d(x)=-\frac{c.x}{a}\,.$$
Now we glue $C_8$ and $C_5$ along $d$.
There is a connected component in the complement of $C$ that contains $(0,0)$. We call the closure of this connected component $D$.
The manifold $D$ (with boundary $C$) obtained from this gluing is a 2 dimensional non-orientable manifold with a connected boundary, therefore it is a M\"{o}bius strip.

Note that since the $d:C_8\to C_5$ is such that 0 is maped to 0,  then there is a branch of the stable manifold of $(0,0)$ and a branch of the unstable manifold of $(0,0)$ that intersect. That is, there is an orbit $\gamma$ such that $$\gamma\subset W^s(0,0)\cap W^u(0,0)\,.$$ We say then that $(0,0)$ has a saddle connection.

 \subsubsection{Completing the vector field to $\mathbb{RP}^2$}

Let us consider a linear vector field in $\RR^2$ with a sink $\omega$, and let us take a neighborhood $u_{\omega}$ in its basin of attraction. We choose a curve in the boundary such that the vector field is  pointing inwards.
Now we can glue $u_{\omega}$ to $D$ Identifying an arc in  the boundary of $u_{\omega}$ with  $C_3$ along a diffeomorphism,  since the vector field $Y$  is pointing outwards at $C_3$.

 Note that the unstable orbit that is not in the saddle connection in $D$ now has its $\omega  $-limit in $\omega$, after the gluing.

 We call the new vector field $Y$ and what remains of the boundary of  $u_{\omega}$, we now call it $C_3'$.

 Analogously we attach a neighborhood $u_{\alpha}$, containing a source $\alpha$
 and glue it to $D$ by identifying a curve in the boundary of   $u_{\alpha}$ witht the segment $C_7$. We call the subset of boundary of $u_{\alpha}$, that was not glued to $D$,  $C_7'$.

Note that the stable orbit  of $(0,0)$ that is not in the saddle connection in $D$ now has its $\alpha$-limit  in $\alpha$.

 We call $D'$ to the region formed by $D$ with $u_{\alpha}$ and $u_{\omega}$ attached.
Since $D'$ is a M\"{o}bius strip, then the complement in
 $\mathbb{RP}^2$ is a disc $R$ having a boundary formed by 4 disjoint curves tangent to the flow ($C_1$, $C_2$, $C_6$ and $C^4$), one curve transverse to the flow and entering $D'$ $C_7'$, and one curve transverse to the flow and exiting  $D'$ $C_3'$. Therefore we can define the flow in the complement of $D'$ in the trivial way by sending the points in  $C_3'$ to  $C_7'$.

 Now we prove Lemma \ref{propY}
\proof
\begin{itemize}
\item Since the original maps are linear, the resulting map after the gluing is also $\mathcal{C}^{\infty}$.
\item The contracting and expanding  Lyapunov values of $Y$ can be taken to be as strong as required
\item As noted above,  one branch of each stable and unstable manifold form a saddle connection $\gamma$ while the others come or go to the sink and source.
\item  The segment, $T_0=C_5$ is a transverse section to $\gamma$ by construction and    is such that:\begin{itemize}
    \item  If $s>0$ $\phi^Y(s,t)$  never touches $T_0$  for any $t>0$ and has $\omega$-limit in $\omega$. And for $t<0$ there exists only one $t_s<0$ such that $\phi^Y(s,t_s)=s'\in T_0$ with $s'<0$. and the $\alpha$-limit of $s$ is $\alpha$.
    \item If $s<0$ $\phi^Y(s,t)$  never touches $T_0$  for any $t<0$ and has $\alpha$-limit in $\alpha$. And for $t>0$ there exists only one $t_s>0$ such that $\phi^Y(s,t_s)=s'\in T_0$ with $s'>0$ and the $\omega$-limit of $s$ is $\omega$.

\end{itemize}
As a consequence of the fact that that $C_8$ was glued to $C_5$ reverting orientation.
\end{itemize}
\endproof

\subsection{Construction of the diffeomorphism $H$}\label{s.conH}

In this subsection we prove the following Lemma from subsection (\ref{s.isotopy}) :
\begin{lemm*}(\ref{l.propH})
Let $\Sigma\subset S^3 \times \mathbb{RP}^2$ be $\sum=S^3\times T$ where $T\subset  \mathbb{RP}^2$ is a uni-dimensional  transverse section to the vector field $Y$ parametrized by  $[-1,1]$.
There exist a $\mathcal{C}^{\infty}$ diffeomorphism isotopic to identity,  $H:\Sigma\to\Sigma$,
$H(x,l)=(r_l(x),\theta_x(l))$
 where $r_l: \sum\to S^3$, $\theta_x:\sum\to T $, with  following properties:
\begin{itemize}

\item $H$ is the identity on the boundary of $\sum$  and  outside of $V_a$ and in $V_r$.
\item The map $r_l(x)$ is the identity for $l=1$ or  $l=-1$, or if $x\in \overline{D_r}$.
\item
The image of $r_l(B_a)=B_r$, and $r_l(p)=p'$ for all $l\in[-1/2,1/2]$.

\item  If $l\in[-1/2,1/2]^c$ then $\theta_x(l)=l$.
\item  If $l\in[0,1/2]$ and $x\notin B_a$ $\theta_x(l)>0$ .
\item  If $l\in[0,1/2]$ and $x\in B_a$ then $-\epsilon\leq\theta_x(l)\leq \epsilon$,

\item  The only point $l$ such that $H(p,l)=(p',0)$, is $l=0$.
\item $H(S^3\times\set{0})\times\set{0}\in\Sigma\times\set{0}$  cuts transversally $S^3\times\set{0}\times\set{0}\in\Sigma\times\set{0}\,.$

\end{itemize}
\end{lemm*}

\proof

Let us consider  a  closed neighborhood of $D_a\cup D_r$ that we call $C$.
Since $B_a$ and $B_r$ are subsets of $S^3$, they are isotopic to each other. Moreover,  we can choose  $C$   so that they are isotopic to each other in $S^3\setminus C$, since $D_a\cup D_r$ does not disconnect $S^3$.
Therefore there is a function $r':S^3\setminus C^{\mathrm{o}} \times[0,1/2]\to S^3$ such that $$r'(x,0)=id(x) \text{ and }r(B_a,1/2)=B_r.$$
We can choose $r'$ so that it is the identity in the boundary of $C$ for all $t\in[0,1/2]$ and such that $r'(p,1/2)=p'$
We can extend now this function to $S^3$ by asking that $r'\mid{D_a\cup D_r}=Id$.
Now $r:S^3\times[-1,1]\to S^3$ is defined by
\begin{equation}
    r(x,l)=\left\{
\begin{array}{ll}
r'(x,l+1), & \mathrm{ } \text{ if }  l\leq -1/2 \\
r'(x,1/2), & \mathrm{  } \text{ if }  -1/2<l\leq 1/2\\
r'(x,l-1), & \mathrm{ } \text{ if }  l>1/2\,.
\end{array}
\right.
\end{equation}
Now we need to construct $\theta:\sum\to[-1,1]$.

We consider a $\mathcal{C}^{\infty}$ bump function $h:[-1,1]\to[-1,1]$,
\begin{itemize}
\item  If  $l\in[-1/2,1/2]^c$ then $h(l)=0$,
\item if   $l\in[-1/2,1/2]$ then $0<h(l)\leq\frac{\epsilon}{2}$,
\item if   $l=0$ then $h(l)=\frac{\epsilon}{4}$,
\item  $\frac{\partial h(l)}{\partial(l)}\mid_{(0)}\neq 0$
\item  $\frac{\partial h(l)}{\partial(l)}< 1$ for all $l$
\end{itemize}
We can also assume that $h$ is sufficiently differentiable.
Let $B_A$ be an arbitrarily small neighborhood of $B_a$.

We consider now a second bump function $g:S^3\to[-1,1]$
\begin{itemize}
\item  If  $x\in B_A^c$ then $g(x)=0$,
\item If  $x\in B_a$ then $\epsilon\leq g(x)\leq\frac{\epsilon}{2}$,
\item  $g(p)=-\frac{\epsilon}{4}$ and $\frac{g(x)}{\partial(v)}\neq 0$ for any given $v$ direction in $S^3$.
\end{itemize}
 We define then  $\theta_x$ as follows:
$$\theta_x(l)=id(l)+h(l)+g(x)\,.$$
Since $\frac{\partial h(l)}{\partial(l)}< 1$ for all $l$, $\theta_p(l)=id(l)+h(l)+g(p)\,$ is an increasing function that is $0 $ only at $l=0$.

Note that the image of the  vectors tangent to the coordinates in $S^3\times \set{0}$, under the differential of $H$, have a non vanishing component in the direction of $T$.
This is our desired function.
\endproof

\subsection{ Construction of the plug}\label{plug}

The aim of this subsection is to  prove the following theorem:

\begin{theo*}[\ref{t.tube}]
There exist a vector field $\chi$ such that its flow  $\phi_{\chi}$  defined  in $S^3$
 has  the following properties:

 There is a region $S^2\times{[-1,1]}\subset S^3$ such that
\begin{itemize}
\item The  vector field $\chi$ is entering at $S^2\times\set{-1}$ and points out at $S^2\times\set{1}$
\item The  vector field $\chi$ is such that the chain recurrent set consists of 2  sources  singularities, $p_1$ and  $p_1'$,  2 sinks singularities, $p_2$ and $p_2'$, and  2 periodic saddles, $p_3$ and $p_3'$.

 \item The intersection of the invariant manifolds of the saddles,  with the boundary of $S^2\times{[-1,1]}$, are disjoint circles that we name as follows:
  \begin{itemize}
  \item $W^{s}(p_3)\cap S^2\times{[-1,1]} = c_0$ in $S^2\times\set{-1}$,
  \item $W^{s}(p'_3)\cap S^2\times{[-1,1]} = c'_0$ in $S^2\times\set{-1}$,
\item $W^{u}(p_3)\cap S^2\times{[-1,1]} =c_1$ in $S^2\times\set{1}$,
\item  $W^{u}(p'_3)\cap S^2\times{[-1,1]} = c'_1$ in $S^2\times\set{1}$.
 \end{itemize}
\item The circle $c_0$ bounds a disc not containing $c'_0$, that we call $D_0$. The circle $c'_0$ bounds a disc containing $c_0$, that we call $D'_0$. And they both bound an open annulus called $A_0$.  Analogously we define $D_1$, $D' _1$ and $A_1$.
\item  The orbit $O(x)$ of a point $x$ in $S^2\times\set{-1}$, crosses  $S^2\times\set{1}$ if and only if $x\in A_0$ and $O(x)\cap S^2\times\set{1}\in A_1$,

\item There is a well defined crossing map $P:A_0\to A_1$. Consider the radial foliation $V_0$ in $A_0$  Then the image of a radial foliation  under $P$ intersect transversally a radial foliation  in $A_1$ and it extends to a foliation in $A_1\cup c_1 \cup c'_1$.
 \end{itemize}
The complement of $S^2\times{[-1,1]}$ in $S^3$ are 2 balls, one in the basin of attraction of a source $r$ (that has $S^2\times\set{-1}$ in the boundary ), and the other in the basin of attraction of a sink $a$.
\end{theo*}

We consider the set $K=\set{(x,y) \,\,\,\text{ tq }\,\,\abs{y}\leq 1\text{ and  } 0\leq x\leq 1}$, and in this set, a flow $\phi_0$ of a vector field $Y_0$ in $\RR^2$ with the following properties:\begin{itemize}
                                        \item The vector field $Y_0$ is Morse-Smale with a source  $p_1=(0,1/2)$, a sink  $p_2=(0,-1/2)$ and a  saddle  $p_3=(1/2,0)$.
                                        \item The vectirfield simetric (in $\RR^2$) with respect to $x$ in a neighborhood of the interval $\set{(0,y) \,\,\,-1\leq y\leq 1}$
                                        \item The saddle $p_3$ is such that  a branch of the unstable manifold intersects the basin of the sink,the other  intersects a corner of $K$.
                                         \item The saddle $p_3$ is such that  a branch of the stable manifold intersects the basin of the source,the other  intersects a corner of $K$.
                                      \end{itemize}

We take a point $q =(1,1-\epsilon)$  for some positive and small $\epsilon$, and  another point of the orbit of $q$ that we call $q'$ with $x$ coordinate $1$.
 We call $K'$ to the "square" delimited by
 \begin{itemize}
 \item the segments $\set{(0,y)\,\,\,tq\,\,\abs{y}\leq 1}$,
 \item the segment $\set{(x,1)\,\,\,tq\,\,0\leq x \leq 1}$,
 \item  the segments $\set{(1,y)\,\,\,tq\,\,1-\epsilon\leq y\leq 1}$,
 \item the vertical segment that joints $q'$ with $(1,-1)$,
 \item  the orbit segment joining $q$ and $q'$,
 \item  the segment $\set{(x,-1)\,\,\,tq\,\,0\leq x \leq 1}$.
 \end{itemize}
The set $K'$ is diffeomorphic to  $$C=\set{(x,y)\,\,\tq \,\, 0\leq x\leq 1 \,\,\text{ y } \, \abs{y}\leq 1}\,$$ where the orbit of $q$ is sent to the segment $[1,y]$ and out of a small neighborhood of this segment everything is left unchanged.
\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.48\linewidth]{flujoplano2.pdf}
\end{center}

\end{figure}

 We call $Y_1$ to the vector field tangent to the flow $\phi_1$ obtained by composing the flow with the diffeomorphism above.

We define a $C^{\infty}$ function $f:\RR\to\RR$ such that:
  \begin{itemize}
    \item $f(0)=0 $
    \item $f(1/2)=1$
    \item $f$ is decreasing  $(1/2,1)$
    \item $f'(x)\neq0$ in $[1/2,1]$
    \item $f(1)=0$.
   \end{itemize}

Now we consider  $C\times S^1$ where if $(x,y,\theta)\in C\times S^1$ then $(x,y)\in C $ with $x\in[0,1]$ and $y\in[-1,1]$.  In $S^1 $ we consider the vector field in $C\times S^1$
that if $$v\in T(x,y)C\,\,\,\text{ then } \chi^{+}=(Y_1(v),f(x))\,.$$
We can also consider in another copy  of $C\times S^1$ the vector field $$\chi^{-}=(Y_1(v),-f(x))\,.$$
 We call $p'_1$, $p'_2$ and $p'_3$ to the source, the sink and the saddle for $\chi^{-}$.
t

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.48\linewidth]{flujoplano.pdf}
\end{center}

\end{figure}
We re write  $C\times S^1$ in Cartesian coordinates as $\mathbb{D}^2\times [-1,1]$ so that the vector field is entering at $\mathbb{D}^2\times\set{1}$, points out at  $\mathbb{D}^2\times\set{-1}$ and is tangent to  $(\partial(\mathbb{D}^2)) \times{[-1,1]}$. We can do this change of coordinates and still have a differentiable vector field because of the symmetry properties of $Y_0$
We paste $2$ copies of  $\mathbb{D}^2\times [-1,1]$ along  $(\partial(\mathbb{D}^2)) \times{[-1,1]}$.
In one copy we have $\chi^{+}$ and in the other we have $\chi^{-}$. Since the vector fields are equal in  $\partial(\mathbb{D}^2) \times{[-1,1]}$, both are $C^{\infty}$ even restricted to the boundary and no orbit crosses  $\partial(\mathbb{D}^2) \times{[-1,1]}$, we can define a gluing map such that  the resulting vector field  $\chi$  defined in  $S^2\times [-1,1]$ is smooth.

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.48\linewidth]{ciltrans.pdf}
\end{center}

\end{figure}

For our convenience, the intersections of the stable and unstable manifolds of the saddle periodic orbits with  $\partial(S^2\times [-1,1])$ are named as follows.
 \begin{itemize}
  \item $W^{s}(p_3)\cap S^2\times{[-1,1]} = c_0$ in $S^2\times{-1}$,
  \item $W^{s}(p'_3)\cap S^2\times{[-1,1]} = c'_0$ in $S^2\times{-1}$,
\item $W^{u}(p_3)\cap S^2\times{[-1,1]} = c_1$ in $S^2\times{1}$.
\item  $W^{u}(p'_3)\cap S^2\times{[-1,1]} = c'_1$ in $S^2\times{1}$.

 \item The intersection of the invariant manifolds of the saddles  with the boundary of $S^2\times{[-1,1]}$ are the disjoint circles, $c_0$,  $c'_0$,  $c_1$, $c'_1$.
 \item The circle $c_0$ bounds a disc not containing $c'_0$, that we call $D_0$. The circle $c'_0$ bounds a disc not containing $c_0$, that we call $D'_0$. And they both bound an annulus called $A_0$  Analogously we define $D_1$, $D' _1$ and $A_1$.
 \end{itemize}

To complete the flow to $S^3$ we add a ball contained in the basin of attraction of a a sink $a$ and a ball contained in the basin of repulsion of a source $r$ in the remaining space . We can do this since  the vector field is entering at  $S^2\times{1}$, and vector field is pointing out at  $S^2\times{-1}$.

 This way, the orbits  behave in one of the following ways (see figure \ref{compflow.f}) :
 \begin{enumerate}
   \item They go from  from $r$ to $a$ and cross the annuli $A_0$ and $A_1$.
  \item They go to the periodic orbit $p'_3$  for the future or the past.
 \item They go for the future to  $p_2$ or $p_2'$,
\item They go for the past to  $p_1$ or  $p_1'$
 \end{enumerate}

\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.55\linewidth]{compleeting.pdf}\label{compflow.f}
\end{center}
\caption{The vector field $\chi$ in $S^3$.}
\end{figure}

The next lemma is to check all the conditions of theorem \ref{t.tube}, except for the transversality condition, that we will check in the next subsection
\begin{lemm}\label{t.tube1}
The vector field $\chi$ defined in $S^2\times [-1,1]$ has the following properties:
\begin{itemize}
\item The  vector field $\chi$ is such that the chain recurrent set consists of 2 sources $p_1$ and  $p_1'$ ,  2 sinks $p_2$ and $p_2'$, and  2 periodic saddles $p_3$ and $p_3'$.

\item  The orbit $O(x)$ of a point $x\in S^2\times\set{-1}$ crosses  $S^2\times\set{1}$ if and only if $x\in A_0$ and $O(x)\cap S^2\times\set{1}\in A_1$
\end{itemize}
\end{lemm}
\proof
The flow $\phi_{Y_1}$ is such that the only chain  recurrent points are the sinks or sources. Therefore all points except for $p_1$,  $p_1'$,   $p_2$, $p_2'$, $p_3$ and $p_3'$ are non-chain recurrent  for  $\chi^{+}$ and  $\chi^{-}$.
For  $\chi^{+}$ over $C\times S^1$   the points of the form $(0,y,\theta)$ have zero $S^1$ coordinate and so $p_1$ and $p_2$ are singularities. Moreover, the eigenvalues of the differential of the vector field at  $p_2$ are one real and negative and one complex with negative real part.
The same reasoning gives us that $p_1$ is a source and for $\chi^{-}$ that $p_1'$ is a source and $p_2'$ is a sink.
For  $\chi^{+}$ the point $p_3$ is periodic with its orbit tangent to the $S^1$ coordinate, therefore since for $Y_1$ the point $p_3$ was a saddle, it is a saddle for $\chi^{+}$  as well. The same reasoning gives us that $p_3'$ is a saddle periodic orbit for $\chi^{-}$.
So for the vector field $\chi$,
by construction, there are no orbits crossing from one copy of  $\mathbb{D}^2\times [-1,1]$ to the other therefore we have that  $\chi$ is such that the chain recurrent set consists of 2 sources $p_1$ and  $p_1'$ ,  2 sinks $p_2$ and $p_2'$, and  2 periodic saddles $p_3$ and $p_3'$.

The second item comes from the fact that all orbits of the points between the stable manifold of $p_3$ and $q$ in the segment $[1,y] $ cross again $[1,y]$ between the unstable manifold of $p_3$ and $q'$ for the flow  $\phi_{Y_0}$.
These are the half open segments that (up to a diffeomorphism) are the base of half of   $A_0$  and  $A_1$ in the product corresponding to  $\chi^{+}$. The other half of   $A_0$  and  $A_1$ that correspond to  $\chi^{-}$ behaves in the same way.
The other orbits of the vector field   $\phi_{Y_0}$. are in the basin of attraction of the sink, or the basin of repulsion of the source or are the singularities themselves. This is also true in the base of the product dynamics of  $\chi^{+}$  and therefore for  $\chi^{+}$ . The dynamics for $\chi^{-}$  is the same.

\endproof

\subsubsection{A radial foliation and the crossing map $P$}

The aim of this subsection is to prove the last part of Theorem \ref{t.tube}.

Since every orbit of the points in $A_0$ cuts $A_1$ at some moment, we define the first return  map $P:A_0\to A_1$. We take polar coordinates in $A_0$ and $A_1$ (that is, we take coordinates in $S^1\times(0,1)$). We write the diffeomorphism $P$ in this coordinates as $$P(\theta,r)=(P_{\theta}(\theta,r),P_{r}(\theta,r))$$
\begin{figure}[htb]
\begin{center}
\includegraphics[width=0.55\linewidth]{acostar.pdf}
\end{center}
\caption{The lift of the  map $P:A_0\to A_1$ to  $\RR\times(0,1)$}
\end{figure}

\begin{lemm}\label{l.transvers}
Let us write  $A_0$ and $A_1$ in coordinates $S^1\times(0,1)$ and
let $P:A_0\to A_1$ be the first return map from $A_0$ to $A_1$, $P(\theta,r)=(P_{\theta}(\theta,r),P_{r}(\theta,r))$, defined by the vector field $\chi$.
If $D P(0,r)=(w,z)$ then $w\neq 0$ for all $r \in (0,1)$.
As a consequence, the image of the segment $(0,r)$ with $r \in (0,1)$,   cuts transversally any segment of the form $(0,r)$ with $r \in (0,1)$ in $A_1$.

\end{lemm}

\proof
 From the construction of $\chi$ there is a circle $ S^1\times q \in A_0$ where   $\chi^{+}$ and  $\chi^{-}$ coincide. We consider new coordinates in  $A_0$ and $A_1$ so that the coordinates of $q $ are now  $(\theta,1/2)$.

Let us first consider a lift of $A_0$, to the strip  $\widehat{A_0}$ parametrized by $\RR\times(0,1)$, we also take the lift of $A_1$, to $\widehat{A_1}$, and the lift of $P$, $\widehat{P}$.
We we may choose  this lifts so that the rotation in the sense of $p_3$ is lifted  as positive.

Let us take a point $x=(\theta,r)$ in  $A_0$. We define the  time that it takes for $x$ to reach $A_1 $ under the flow of  $\chi$ as $T_x$.

Suppose that $r<1/2$ Recall that the vector field  is $\chi=(Y_1,Y_2)$ were  $$Y_2(\theta)=f(g(r)) \;d\theta\,,$$ and  therefore $$\frac{\partial P_{\theta}(\theta,r)}{\partial r}=T_x f(g(r))'g(r)'\,,$$
 and therefore non vanishing.
 Suppose that $r>1/2$,  then the vector field $Y_3$ is defined as $$Y_3(\theta)=-f(g(r)) \;d\theta\,,$$ and  therefore $$\frac{\partial P_{\theta}(\theta,r)}{\partial r}=T_xf(g(r))'g(r)'\,,$$
 and therefore non vanishing.

 At $r=1/2$ since the lateral derivatives are not 0 and the function is smooth then.

 $$\frac{\partial P_{\theta}(\theta,r)}{\partial r}\neq0\,.$$

Note that from the construction of the flow $T_x$ goes to infinity as $r$ goes to $0$ or $1$
\endproof

config: arxiv


TITLE: How to complete this simple example of the Vitali-Hahn-Saks Theorem?
QUESTION [4 upvotes]: I'm studying the Vitali-Hahn-Saks Theorem and I need some help with an example in which the conclusion of the theorem fails.
Theorem. Let $(\Omega,\mathcal{F}, P)$ be a finite measure space, and let $(P_n)$ a sequence of finite measures on $(\Omega, \mathcal{F})$ each of which is absolutely continuous with respect to $P$. Suppose the sequence $(P_n(\Omega))$ is bounded. If $(P_n)$ converges setwise to the set function $P_\infty$, then $(P_n)$ is uniformly absolutely continuous with respect to $P$ ($\forall \epsilon > 0, \exists \delta>0, \forall n, \forall A \in \mathcal{F}, P(A) < \delta$ implies $P_n(A) < \epsilon$). Moreover, $P_\infty$ is a finite measure that is absolutely continuous with respect to $P$.
I thought I'd cook up a simple example where uniform absolute continuity fails in order to see how setwise convergence fails. Start with a countable probability space given by $P(\omega_n) = 2^{-n}$. Let the events $E_n = \Omega - \cup_{i=1}^n \{\omega_i\}$. And let $P_n = P(\cdot \mid E_n)$. Note that $\inf_n\{P(E_n) \}=0$.
The sequence $(P_n)$ is not uniformly absolutely continuous with respect to $P$. To see that let $\epsilon = 1/2$, and let $\delta > 0$ be given. For large enough $n$, $P(E_n) < \delta$ but $P_n(E_n) = 1 > \epsilon.$

Question. How to see that $(P_n)$ does not converge setwise? I've played around with a few example events $A$, but $(P_n(A))$ always converges. Hints are appreciated.

REPLY [1 votes]: Here are some hints about what kind of set $A$ you are looking for:
If $A$ is finite, then eventually $A\cap E_n = \emptyset$, so
$$
P_n(A)=\frac{P(A\cap E_n)}{P(E_n)} \longrightarrow 0
$$
Thus we are looking for infinite sets. Moreover, if we eventually have $E_n \subset A$ then $A \cap E_n = E_n$ so
$$
P_n(A) = \frac{P(A \cap E_n)}{P(E_n)} = \frac{P(E_n)}{P(E_n)} \longrightarrow 1
$$
Thus we are looking for a set $A$ that has infinitely many "holes" (more precisely, for any $N$ we can find a $n >N$ such that $\omega_n$ is not in $A$).
Can you think of a very simple set $A$ with these properties?

Added later:
Elaborating on the hint given above, one of the simplest set to satisfy the properties we want is the set of $\omega_i$ with even index. That is, we take
$$
A = \{ \omega_{2 i} : i \geq 1 \}
$$
In this case, after some calculation we find that $P_n(A) = 1/3$ when $n$ is even and $P_n(A) = 2/3$ when $n$ is odd. Consequently, the sequence $P_n(A)$ does not converge.