Given a function $f:X\rightarrow Y$ where $A$ and $B$ are subsets of $X$, prove that $A\subset B \Rightarrow f(A)\subset f(B)$

Question

I know this question has already been posted here $A$ subset $$ implies $f(A)$ subset $f()$. However, I would like to know if my approach is just as fine.

Let $f:X\rightarrow Y$ be a mapping from $X$ into $Y$. Show that if $A$ and $B$ are subsets of $X$, then

$$(A\subset B)\Rightarrow (f(A)\subset f(B))$$

MY ATTEMPT

Indeed, due to the hypothesis, which is equivalent to $A\cap B = A$, one has \begin{align*} f(A) & = \{y\in Y \mid\exists x\,(x\in A)\wedge(y = f(x))\} = \{y\in Y \mid\exists x\,(x\in A\cap B)\wedge(y = f(x))\} \\\\ & = \{y\in Y \mid\exists x\,(x\in A)\wedge(x\in B)\wedge(y = f(x))\}\\\\ & = \{y\in Y \mid\exists x\,(x\in A)\wedge(y = f(x))\}\cap\{y\in Y \mid \exists x\,(x\in B)\wedge(y = f(x))\}\\\\ & = f(A)\cap f(B) \subset f(B) \end{align*}

Am I on the right track? Any help is appreciated.

Answer 1

If y in f(A), then exists a in A with y = f(a).
Thus exists a in B with y = f(x); y in f(A).

Answer 2

One flaw in your proof: The statement $$ \{y\in Y \mid\exists x\,(x\in A)\wedge(x\in B)\wedge(y = f(x))\}\\ = \{y\in Y \mid\exists x\,(x\in A)\wedge(y = f(x))\}\cap\{y\in Y \mid \exists x\,(x\in B)\wedge(y = f(x))\} $$ is incorrect. There is only set inclusion ($\subset$) between the left and right sides, not equality. Fortunately your proof goes through with this correction.

The bulk of your argument amounts to a proof that $$ f(A\cap B)\subset f(A)\cap f(B),\tag a $$ which is part (a) of the question that you've linked. Having proven (a), you can apply it to shorten your argument for (b):

If $A\subset B$, then $A=A\cap B$ so $f(A)=f(A\cap B)\subset f(A)\cap f(B)\subset f(B)$.