Is the first part of my set theory proof correct?

Question

I'm stuck with the following problem and I'm wanting to know if my proof is strong enough; and if not, what I can do to make it more rigorous. Set Theory proofs are really confusing me so apologies if I've made any steps in error.

Proof:

Take any $x,y$ such that $x \in S$ or $x \in T$, with $y = f(x), y \in B$.

Therefore, $x$ must lie in the union of $S \bigcup T$ by definition of the union; therefore, $y \in f(S \bigcup T)$ for all $x$.

Also, $y\in f(S)$ for all $x \in S$; further $y \in f(T)$ for all $x\in T$. Therefore, $y \in f(S) \subseteq f(S \bigcup T)$ and $y \in f(T) \subseteq f(S \bigcup T)$

Thus, $y \in f(S) \bigcup f(T)$ for all $x$. Therefore, $y \in f(S \bigcup T) $ and $y \in f(S) \bigcup f(T)$; therefore these two sets are equal.

Is this correct? In order to prove equality I need to go the other way; but I am unsure how to start this. Can I have someone review this and point in me the right direction to finish my proof? I understand I'm going to have to show for any $y \in f(S) \bigcup f(T)$ it also exists in $f(S\bigcup T)$ but is it really this simple?

Thank you for all responses.

Answer 1

Your idea is correct, but your language is not. In order to prove that $f(S\cup T)\subset f(S)\cup f(T)$, take $x\in S\cup T$. Then you should prove that $f(x)\in f(S)\cup f(T)$, but that's easy: if $x\in S$, then $f(x)\in f(S)$ and if $x\in T$, then $f(x)\in f(T)$.

In order to prove the reverse inclusion. take $y\in f(S)\cup f(T)$. Then $y\in f(S)$ or $y\in f(T)$. If $y\in f(S)$, then $y=f(x)$ for some $x\in S$ and if $y\in f(T)$, then $y)f(x)$ for some $x\in T$. SO, $x\in S\cup T$ and therefore $y\in f(S\cup T)$.

Answer 2

In the first part you actually prove that $x\in S\cup T\implies f(x)\in f(S\cup T)$ which is a nothing more than a consequence of the definition of $f(S\cup T)$.

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Let $y\in f(S\cup T)$. Then by definition $y=f(x)$ for some $x\in S\cup T$. If $x\in S$ then we have $y=f(x)\in f(S)$ and also - because $f(S)$ is a subset of $f(S)\cup f(T)$ - we have $y\in f(S)\cup f(T)$. If $x\in T$ then likewise we find $y\in f(S)\cup f(T)$. Proved is now $y\in f(S\cup T)\implies y\in f(S)\cup f(T)$ and this enough to conclude that $f(S\cup T)\subseteq f(S)\cup f(T)$.

From $S\subseteq S\cup T$ it follows directly that $f(S)\subseteq f(S\cup T)$. Likewise we find that $f(T)\subseteq f(S\cup T)$, and these two facts allow us to conclude that $f(S)\cup f(T)\subseteq f(S\cup T)$.

Now we proved both sides and we are ready.