Equivalence Relations Proof 1234

Question

Let $Q=\{1,2,3,4,5,6,7,\dotsc,26\}$ and let $P$ be the non-empty subsets of $Q$. For $A \in P$ and $B \in P$, $A\sim B$ iff there is a bijection $f\colon A \to B$.

How do I prove that this is an equivalence relation?

Also, I know that...

$i\colon A\to A$ defined by $i(x)=x$ is a bijection. $f\colon A\to B$ is a bijection, then $f^{-1}\colon B\to A$ is well defined and a bijection. $f\colon C\to B$ and $g\colon B\to A$ are bijections when $g \circ f\colon C\to A$ is also a bijection.

Answer 1

Hint: What conditions must a relation satisfy in order for it to be an equivalence relation? Find out what they are and then show that the relation in question satisfies them.

Answer 2

Here are my suggestions for rewriting your last paragraph (which really conatins all the mathematics it needs) to a full proof. Note that this is far from the only way to write it, but it should at least be an improvement.

From the original paragraph:

$i\colon A\to A$ defined by $i(x)=x$ is a bijection.

This is meant to prove reflexivity, so let's say that explicitly (actually, that goes for all three properties). Mention that you want to prove that $A\sim A$, and that this means that you need to find a bijection $i:A\to A$. For instance, if you change the above line to

Given an arbitrary $A\in P$, we have a bijection $i:A\to A$ given by $i(x) = x$. Therefore $A\sim A$, and because $A$ was arbitrary, we have that $\sim$ is reflexive.

then it is a lot easier to follow your line of thought.

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Next line:

$f\colon A\to B$ is a bijection, then $f^{-1}\colon B\to A$ is well defined and a bijection.

Again, let's mention explicitly that you're after symmetry here. I suggest something like

Given arbitrary $A, B\in P$ with $A\sim B$, then by definition there is a bijection $f:A\to B$. Because $f$ is a bijection, it has an inverse $f^{-1}:B\to A$, which is also a bijection. This implies $B\sim A$, so $\sim$ is symmetric.

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Lastly, we have transitivity:

$f\colon C\to B$ and $g\colon B\to A$ are bijections when $g \circ f\colon C\to A$ is also a bijection.

As well as suggesting we mention transitivity and $\sim$ explicitly, you choose your function directions a bit unconventionally. I therefore suggest changing the order of things a bit, just to conform a bit more to the expectations of someone reading this. Perhaps something along these lines?

Given arbitrary $A, B, C\in P$ with $A\sim B$ and $B\sim C$, we have by definition bijections $f:A\to B$ and $g:B\to C$. Their composition $g\circ f$ is a bijection from $A$ to $C$, which means that $A\sim C$. Therefore $\sim$ is transitive.

Lastly, when you write these together, I would personally write them in some sort of list form, either a bulleted list, or each of the paragraphs separately with "Reflexivity", "Symmetry" and "Transitivity" being underlined titles (perhaps the latter is better for hand written text). Anything that makes what you're doing more visible is a good thing.