TITLE: Let $ V $ be open in a metric space $ (X,d) $. Show that, for all $ A⊂X $, $ V∩\overline{A}⊂\overline{A∩V} $.
QUESTION [1 upvotes]: Definition: If $ X $ is a metric space, $ E \subset X $ and $ E '$ represents the set of all limit points of $ E $ in $ X $, then let us denote the closure of $ E $ as $ \overline {E} = E \cup E '. $

Let $ V $ be an open in a metric space $ (X,d) $. Show that for all $ A⊂X $ we have $ V∩\overline{A}⊂\overline{A∩V} $.
My attempt.
Let $ x \in V \cap \overline {A} $, then $ x \in V $ and $ x \in \overline {A} $. Let us consider two cases:

If $x\in A$ and $x\in V$, then $x\in V\cap A\subset \overline{A\cap V}$. Since for any set $ A $ we have $ A \subset \overline {A} $.
If $x\in A'$ and $x\in V$. Since $ x \in A '$, there exists $ q \neq x $ such that $ q \in A $. Since $ V $ is open, then all its points are limits, that is, $ q \in V $. Therefore, $ q \in V \cap A $, that is $ x \in \overline {V \cap A} $.

REPLY [3 votes]: No, the proof is not correct.
The division into cases is good and the proof for the first case is OK.
If $x\in A'$ and $x\in V$, then you want to show that $x\in(V\cap A)'$.
Take a neighborhood $U$ of $x$. Then also $U\cap V$ is a neighborhood of $x$ and, by definition of limit point, there is $y\in (U\cap V)\cap A=U\cap(V\cap A)$, $y\ne x$. Thus $y\in V\cap A$ and this proves that $x\in (V\cap A)'$. Note that, by definition, $(V\cap A)'\subseteq\overline{V\cap A}$, so the proof is complete.


TITLE: Continuous maps from products of topological spaces
QUESTION [7 upvotes]: Let $X,Y,Z$ be topological spaces. It is well-known that if $F:X\times Y\to Z$ is a continuous map, we can define a map $$\overline{F}:X\to C(Y,Z) \\\overline{F}(x)(y)=F(x,y)$$ where $C(Y,Z)$ is the topological space of all continuous maps from $Y$ to $Z$ equipped with the compact-open topology, and this induced map is continuous.
If $Y$ is locally compact Hausdorff, the converse holds, i.e. if a function $\overline{F}:X\to C(Y,Z)$ is continuous, and related to $F$ by the equation above, then $F$ is also continuous. This converse seems a bit unsatisfying, so I was wondering:

Does the converse also hold under some weaker assumptions? What is the best known theorem in this regard?

Also, I suspect the problem might be with the choice of topology on $C(Y,Z)$. As nice as the compact-open topology might be, I imagine there might be some other (more or less natural) choice of topology availible for which the continuity of $F$ and continuity of $\overline{F}$ would be equivalent.

Does there exist such a topology?

REPLY [4 votes]: Looking at the paper referenced in the Mathoverflow question, the first line of the abstract states:

It is well-known that a Hausdorﬀ space is exponentiable if and only if
  it is locally compact, and that in this case the exponential topology is the
  compact-open topology.