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Can somebody prove For all subsets $A$ and $B$ of $X$, $f(A \cup B) = f(A) \cup f(B)$ ?

I believe that it is true, and here is my proof. If somebody sees something I did wrong, can you please explain?

Take any $y∈F (A∪B).$ This means $\exists x∈A∪B : f(x)=y.$ Since $x∈A∪B,$ then $x∈A$ or $x∈B,$ so $y=f(x)∈f(A)$ or $y=f(x)∈f(B),$ so $y∈f(A)∪f(B).$ Thus $f(A)∪f(B)⊆f(A∪B).$

Take any $y∈f(A)∪f(B),$ so $y∈f(A)$ or $y∈f(B).$ Assume $y∈f(A).$ This means $\exists x∈A : y=f(x).$ Now $A ⊆A∪B,$ so $x∈A∪B,$ so $y=f(x)∈f(A∪B).$ This shows $f(A)∪f(B)⊆f(A∪B).$

David K
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User
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    Your proof is correct. –  Apr 28 '15 at 04:33
  • Thank you! Can you tell me why proving that f(A)∪f(B)⊆f(A∪B) is sufficient? I would have thought I would have to prove that f(A)∪f(B)=f(A∪B), but I guess proving the subset is adequate, though I am not entirely sure why this worked. – User Apr 28 '15 at 04:35
  • Your proof is fine, though at this level they may want you to explicitly say "without loss of generality" regarding your assumption that y is in f(A). – Jonathan Hebert Apr 28 '15 at 04:36
  • Oh I just spotted a mistake (which might explain your doubt): On the last line of the first paragraph you should have $f(A\cup B) \subset f(A) \cup f(B)$. –  Apr 28 '15 at 04:36
  • @OmarN you did prove both containments. – Jonathan Hebert Apr 28 '15 at 04:37
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    You can find links to several posts about this result here. – Martin Sleziak Apr 28 '15 at 05:09

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Looks great. One thing to point out in the second part: you assume that $y \in f(A)$. Technically, you also need to show that the other case works as well. That is, you technically need to show that even if $y \in f(B)$, we still arrive at the same conclusion that $y \in f(A \cup B)$. The proof is exactly the same as what you did however, and so you might even get away with saying something like: "without loss of generality, we may assume that $y \in f(A)$" or "a similar argument holds for the case when $y \in f(B)$".

Adriano
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As John notes, your proof is correct, but I think it could be written more effectively. As you know, to prove $f(A\cup B)=f(A)\cup f(B)$, we must prove that $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A\cup B)\supseteq f(A)\cup f(B)$ (i.e., $f(A)\cup f(B)\subseteq f(A\cup B)$).

$(\subseteq)$: Suppose $x\in f(A\cup B)$. Thus, $x=f(y)$ for some $y\in A\cup B$. Either $y\in A$, in which case $x\in f(A)$, or $y\in B$, in which case $x\in f(B)$. Thus, in either case, $x\in f(A)\cup f(B)$. This shows that $f(A\cup B)\subseteq f(A)\cup f(B)$.

$(\supseteq)$: Suppose $x\in f(A)\cup f(B)$. Then either $x\in f(A)$ or $x\in f(B)$. This means either that $x=f(y)$ for some $y\in A$ or that $x=f(y)$ for some $y\in B$. In either case, $x=f(y)$ for some $y\in A\cup B$, so $x\in f(A\cup B)$. This shows that $f(A)\cup f(B)\subseteq f(A\cup B)$.

We have shown that $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A)\cup f(B)\subseteq f(A\cup B)$. By mutual subset inclusion, $f(A\cup B)=f(A)\cup f(B)$. $\blacksquare$

Daniel W. Farlow
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  • Thank you! Though can you tell me why to prove $f(A\cup B)=f(A)\cup f(B)$, we must prove that $f(A\cup B)\subseteq f(A)\cup f(B)$ and $f(A\cup B)\supseteq f(A)\cup f(B)$ (i.e., $f(A)\cup f(B)\subseteq f(A\cup B)$)? I am confused as to why to prove $f(A\cup B)=f(A)\cup f(B)$ , we don't try to prove $f(A\cup B)=f(A)\cup f(B)$ – User Apr 28 '15 at 05:07
  • @OmarN Showing mutual subset inclusion to show that two sets are equal is a standard practice unless you can get away with doing some basic set algebra as in this answer. – Daniel W. Farlow Apr 28 '15 at 05:09