TITLE: Why is $dS=\frac{dQ}{T}$?
QUESTION [0 upvotes]: From the sources I've read it I found that it was "found by clausius" but how did he find it? What was the motivation?
I know of the clausius inequality and the fact that summing up the quantity $\frac{dQ}{T}$ over the whole cycle of a heat engine is equal to or less than zero always. However why is that for a reversible process we can write a state function for it??

REPLY [2 votes]: This is a consequence of the Clausius theorem. The background to it is explained in this Wikipedia article.
We know that in a heat engine that goes through a cycle and returns to its original state, we have
$\oint \frac 1 T dQ \le 0$
with equality only if the cycle is reversible. This tells us that if we move from state $A$ to state $B$ in a reversible way then the quantity
$\int_A^B \frac 1 T dQ$
is always the same, regardless of the actual path taken from $A$ to $B$ (as long as it is reversible). And this in turn tells us that there is a quantity
$S = \int \frac 1 T dQ$
(defined up to an additive constant) which is a function of the state of a system alone (i.e. not dependent of the history of that system) as long as the system only undergoes reversible changes. This quantity is what we call entropy.


TITLE: $\bigcup X$ finite implies $\mathcal P(X)$ is finite.
QUESTION [5 upvotes]: Can anyone help with this past paper question from a Set Theory exam.

Prove that, for all sets $X$, $\bigcup X$ finite implies $\mathcal P(X)$ finite.

I am using the Kuratowski definition of finiteness, ie A is finite if every Kuratowski inductive set for A contains A.
Thanks in advance!

REPLY [0 votes]: The key facts are:

If $B$ is finite and $A\subseteq B$, then $A$ is finite; $\quad(1)$
If $A$ is finite, then $\mathcal{P}A$ is finite. $\quad(2)$

Notice that $A\subseteq\mathcal{P}(\bigcup A)$. If this is not immediately obvious: fix $a\in A$, then every $\gamma\in a$ must also be in $\bigcup A$ by definition of the union; it is then clear that $$a=\{\gamma\in\bigcup A:\gamma\in a\}\in\mathcal{P}(\bigcup A).$$
Hence, by $(1)$, if $\mathcal{P}(\bigcup A)$ is finite, $A$ is finite.$\quad(3)$

Putting these facts together:
$$\bigcup X\text{ finite}\quad\stackrel{(2)}{\Rightarrow}\quad\mathcal{P}(\bigcup X)\text{ finite}\quad\stackrel{(3)}{\Rightarrow}\quad X\text{ finite}\quad\stackrel{(2)}{\Rightarrow}\quad \mathcal{P}X\text{ finite}.$$


TITLE: Calculate $\int \frac{r \: dr}{\sqrt{r^4 l^{-2}+ r^2(1 + a^2 l^{-2}) - 2 m r + a^2}}$
QUESTION [0 upvotes]: How to solve this integral?
$\displaystyle \int \frac{r \: dr}{\sqrt{r^4 l^{-2}+ r^2(1 + a^2 l^{-2}) - 2 m r + a^2}}$
where, $a,l,m \in \mathbb{R}$ and $m > 0$.
I tried putting this in Mathematica for some parameter values, and it gave some complicated function of elliptic integrals.

REPLY [0 votes]: $$I=\int\frac{rdr}{\sqrt{c_1r^4+c_2r^2+c_3r+c_4}}$$
where:
$c_1=l^{-2}$, $c_2=1+a^2l^{-2}$, $c_3=-2m$ and $c_4=a^2$
This would be my first try but unlike a quadratic, this quartic does not seem to have a general solution


TITLE: Element of order 6 in A4
QUESTION [2 upvotes]: How do I show that A4 cannot have an element of order 6?

I'm not sure how to go about this. So, we would have to have the element either as (6 cycle)(1 cycle), (1 cycle)(6 cycle), (3 cycle)(2 cycle), and (2 cycle)(3 cycle). The first two are not possible because we only have 4 elements to per-mutate. The other two would lead it to be an odd permutation correct? How do I express this in a more formal way?

REPLY [3 votes]: Take $\sigma \in \Sigma_4$ to be an element of order $6$. Decompose $\sigma$ into a product of cycles, remembering that the order of $\sigma$ is the l.c.m. of the order of its component cycles. Consider the possibilities and show that none of these gives rise to an element in $A_4$.


TITLE: Question about proof of Heine-Cantor (i.e. compact and continuous implies uniform continuous)
QUESTION [0 upvotes]: If anyone has seen the wikipedia page for the Heine-Cantor theorem, I find something off about the proof it presents. It would be incredibly tedious to write it all out here because it's pretty involved, but you can find it at here.
At a certain point in the proof, after $\delta$ has been defined, the author goes on to say that for all $x,y$ in the domain $M$, $d(x,y)<\delta$. Can anyone corroborate this?

REPLY [0 votes]: That's not what it says: suppose that $d_X(x,y) < \delta$ for two arbitrary points of $X$. It might have been less confusing for you if the proof had said: "let $x,y$ be two arbitrary points of $X$ such that $d_X(x,y) < \delta$" instead:
Reason: we want to show
$$\forall x,y \in X: d_X(x,y) < \delta \implies d_Y(f(x),f(y)) < \varepsilon$$
so we start by assuming we have two points in $X$ that are distance $< \delta$ apart, and these must be arbitrary points with that property, it doesn't mean that all points of $X$ are $< \delta$ apart, but if we happen to have any two that are, we must show that the distance of their images is $< \varepsilon$.