So I have a proof that I have written of $X\subset Y \Rightarrow f(X)\subset f(Y)$ but it is slightly different than the one presented in this questions accepted answer.
The difference is subtle so apologies if this turns out to be a non-issue. However, I know--especially in proofs--that details matter.
Here is my proof:
Suppose $X\subset{Y}$
let $a\in X$
Then $a\in Y$ by def of $X\subset{Y}$
Define $b=f(a)$ (!!Unsure whether this is a valid step!!)
Then $b\in f(X)$ by definition of $f(X)=\left \{ y\in Y\,|\, y=f(x)\, for\, some\,\, x\in Y \right \}$
Similiarly since $a\in Y$, $b=f(a)$ also gives us $b\in f(Y)$.
Therefore $A\subset B \Rightarrow f(A)\subset f(B)$
The accepted answer in the linked question makes the assumption $b\in f(X)$ whereas I use;
Define $b=f(a)$
Is my step a somewhat of a weaker assumption since this is was derived from the hypothesis? or is it entirely equivalent?
To be clear--I am not disputing the correctness of the proof in the linked accepted answer and in hindsight, it is more clear than my own proof--my concern is whether or not my step of defining $b=f(a)$ was justified or otherwise equivalent to the assumption $b\in f(X)$ used in the linked answers proof. If not please explain when defining as I have done is justified. I have seen this move/step used in proofs that I study and it is not entirely clear to me when this can be done.
I suspect defining as I have done is nothing more than just an assumption, yet since it is cast in terms of existing symbols I cannot help but wonder if it may be more than an assumption and actually a tautology (and therefore a true statement).
UPDATE
Just for clarification for anyone who stumbles upon this...
The step
let $a\in X$
was the mistake. My proof shows rather that $(x\in X \land X\subset Y) \rightarrow (f(X) \subset f(Y))$ this is because the assumption $x\in X$ is never "discharged".
using the assumption $b\in f(x)$ works to show $f(X)\subset f(Y)$ because it does eventually get discharged. which would then lead to the result that was desired $X\subset Y \rightarrow f(X)\subset f(Y)$
The "definition" itself is fine; you've picked out an element $a$, so there must exist some element $f(a)$ (by the definition of what a function does), and you're just giving that element a new name -- $b$.
Your proof is flawed, however, because you fail to show that for every $b \in f(X)$, $b \in f(Y)$. You only show that for some particular $b$ -- the $b = f(a)$ you have singled out. Now since $a$ is arbitrary, $f(a)$ is also arbitrary, and in fact any $b \in f(X)$ can be expressed as $b = f(a)$ for some $a$... but you should spell all this out.
You can usually (with some practice) figure out what the skeleton of your proof should look like. You want to prove $f(X) \subset f(Y)$, which by definition means $b \in f(X) \implies b \in f(Y)$; thus you should be assuming $b \in f(X)$. Your proof doesn't make that assumption, so it's not clear how it proves what it sets out to prove.