Prove that a relation R on set A is antisymmetric if and only if $R \cap R^{-1} \subseteq \{(a,a):a \in A\}$. Can someone check to see if my proof is correct? If it actually is correct, can someone tell me how to be less verbose and "make it mathy and less wordy" for my backwards implication $(\Leftarrow)$ portion of the proof? Here's the problem:
Prove that a relation $R$ on set $A$ is antisymmetric if and only if $R \cap R^{-1} \subseteq \{(a,a) : a \in A\}$.
My proof attempt:
$(\Rightarrow)$ Let $R$ be an antisymmetric relation on $A$. Suppose $(a,b),(b,a) \in R$ where $a, b \in A$.
Then by definition of antisymmetric it must the case that $a=b$, so $(a,a) \in R \Rightarrow (a,a) \in R^{-1}$
$\Rightarrow (a,a) \in R\cap R^{-1} \Rightarrow (a,a) \in \{(a,a) : a \in A\}$.
Since $a, b$ were arbitrary elements, $R \cap R^{-1} \subseteq \{(a,a) : a \in A\}$
$(\Leftarrow)$ Let $R\cap R^{-1} \subseteq \{(a,a) : a\in A\}$. To be a subset of $\{(a,a) : a \in A\}$, $R\cap R^{-1}$ must be a set of ordered pairs where the 1st and 2nd terms of the ordered pairs are equal to a single element from $A$. Consider then, any arbitrary element of A, say $a \in A$. Then $(a,a) \in A \times A$ and $(a,a) \in R \cap R^{-1}$. By definition of intersection, $(a,a) \in R$ as well. These results hold for all elements of $A$, so $R \subseteq A\times A$. Thus, $R$ is a relation on $A$ by definition.
To say $R$ is an antisymmetric relation on $A$ is vacuously true since there will never be an ordered pair of the form $(x, y)$ or $(y,x)$ in $R$, unless, of course, $x = y$.
Since  $R\cap R^{-1} \subseteq \{(a,a) : a\in A\} \Rightarrow$ the relation $R$ on set $A$ is antisymmetric
And since the relation $R$ on set $A$ is antisymmetric $\Rightarrow R\cap R^{-1} \subseteq \{(a,a) : a\in A\}$
Therefore the relation $R$ on set $A$ is antisymmetric $\Leftrightarrow$ $R \cap R^{-1} \subseteq \{(a,a) : a \in A\}$.

SECOND ATTEMPT:
Thanks for the comments. Is this an "improvement" for my forward implication?
$(\Rightarrow)$ We are given the relation $R$ on $A$ is antisymmetric. WTS that $R \cap R^{-1} \subseteq \{(a,a):a\in A\}$.
Let $a,b \in A$ be arbitrary elements such that $(a,b) \in R\cap R^{-1}$. Then $(a,b) \in R$ and $(a,b) \in R^{-1}$. If $(a,b) \in R^{-1}$ then $(b,a) \in R$. By def of antisymmetric, since we have $(a,b)$ and $(b,a)$ in $R$, it must be the case that $a = b$, that is, $(a,a) \in R$. Then $(a,a) \in R^{-1}$ as well, which means $(a,a) \in R \cap R^{-1}$.
Thus, $(a,b) \in R\cap R^{-1} \Rightarrow (a,a) \in R\cap R^{-1}$, so $R\cap R^{-1} \subseteq \{(a,a): a \in A\}$
Which as you said, is not true since we could have the case where $R = \emptyset$ which is antisymmetric, and any non-empty $A$ like $a \in A$ then $(a,a) \notin R \cap R^{-1}$... how do I get around to showing this result holds for all relations R on A?
 A: Both directions of your proof need improvement. One general theme that goes through both proofs is that... they are simply not proofs. You write sentences containing claims, but you make the justification for those claims very vague, and sometimes non existent. In general, the proofs require a significant rewrite, where I suggest you focus on the following:

*

*Make it clear what your premises are.

*Explain, in the beginning, what you want the conclusion to be.

*Then, make each statement in such a way that it is clear that it either follows from previous statements or from the premises.

*Use standard mathematical wording, such as "Let $x\in X$ be arbitrary". This allows you to (later on) easily draw conclusions, since, if you start with $x\in X$ being arbitrary, and you prove that $P(x)$ is true, then you can conclude that $\forall x\in X: P(x)$ is also true.

To go into details about what is wrong with your proof...

The first half of the proof is confusing.
You need to prove that if $R$ is antisymmetric, then $R\cap R^{-1}\subset\{(a,a):a\in A\}$.
In order to do that, you need to take an arbitrary element of $x\in R\cap R^{-1}$, and prove that $x\in\{(a,a), a\in A\}$. I suggest you rewrite your proof to make that more clear.

The second half of the proof is even worse. Again, what you should do is you should prove that if $R\cap R^{-1} \subseteq\{(a,a): a\in A\}$, then $R$ is antisymmetric.
You prove that by taking an arbitrary pair $a,b\in A$, and showing that if whe know that $(a,b)\in A$ and $(b,a)\in A$, then we can conclude that $a=b$. You did not such thing in your proof.
In fact, you also made a pretty big error where you wrote something that is just not true. In particular, this:

Consider then, any arbitrary element of A, say $a \in A$. Then $(a,a) \in A \times A$ and $(a,a) \in R \cap R^{-1}$

This is not true. Yes, if $a\in A$, you can conclude that $(a,a)\in A\times A$. But you cannot conclude that $(a,a)\in R\cap R^{-1}$. That is just not true. For example, if $R=\emptyset$ (which is an antisymmetric relation), and if $A$ is nonempty, then $a\in A\implies (a,a)\in R\cap R^{-1}$ is most certainly not true.
