[Math] Prove the following statement: If A is any set, then $A \subseteq A$

Question

I'm doing some practice problems and I'm wondering if I got this right. I think this is a very short proof, but I'm not sure.

Given:

P: A is any set

Q: $A \subseteq A$

We have a $P \rightarrow Q$ statement.

Definition 3.1.2 states that we let A and B be sets. Then A is a subset of B, written $A \subseteq B$, when the statement $(\forall x)[x \in A \rightarrow \in B ]$ is true.

Since A is any set, then by Definition 3.1.2, $(\forall x)[x \in A \rightarrow \in A ]$.

Therefore, $A \subseteq A$

I still remember the truth table for $P \rightarrow Q$. Both statements of $P$ and $Q$ had to be true for $P \rightarrow Q$ to be true.
So, if A is any subset, then $A \subseteq A$

If $P$ was false and $Q$ was false, the $P \rightarrow Q$ would be true.
If A isn't any subset, then A is not a subset of A.

If A is a subset, then A is not a subset of A is the only false statement for $P \rightarrow Q$ because $P$ is true, but $Q$ is false.

Answer 1

Looks about right, but there's a ton of fluff and stuff and oh man all you need is this:

Definition: $A \subseteq B$ iff $\forall a\in A, a \in B$.

As $\forall a \in A, a \in A$, we have $A \subseteq A$.