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I've been working these set theory problems for homework and I think I'm getting the gist of them, but there seem to be little nuances that are giving me issues. For example, as mentioned in the title i'm trying to prove $$f[A\cap{B}]\subseteq{f[A]}\cap{f[B]}$$ So I say let $f$ be a function on sets $A$ and $B$ and let $x\in{A\cap{B}}$. Then $f(x)\in{f[A\cap{B}]}$. But $$f[A\cap{B}]=f[\{x|x\in{A} \text{ and } x\in{B}\}=\{f(x)|x\in{A}\text{ and }x\in{B}\}=f[A]\cap{f[B]}$$ Hence, $f(x)\in{f[A]\cap{f[B]}}$, and therefore, $f[A\cap{B}]\subset{f[A]}\cap{f[B]}$.

I believe this is right, but the remainder of the question asks me to show that proper inclusion can occur. I can't use the iff scenario with because that would show equality not the proper subset case. What am i missing?

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marked as duplicate by user63181, Dan Rust, Norbert, TZakrevskiy, ronno Dec 20 '13 at 13:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Hint: If $f:\mathbb N\to\mathbb N$ is a constant function what can you say about $f[A\cap B]$, where $A$ are the even numbers and $B$ the odd numbers.

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  • $\begingroup$ That is exactly what I am looking for! So all you need is one example to imply a proper subset... $\endgroup$ – Iceman Sep 17 '13 at 21:31
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Note that $y\in f[A]\cap f[B]$ doesn't imply that there is an $x$ in $A\cap B$ with $f(x)=y$. It simply means that there are $a\in A, b\in B$ with $f(a)=f(b)=y$.
E.g. if $A=\{1,2\}, B=\{2,3\}$ and $f:\{1,2,3\}\to \{1,2\}:1\mapsto 1, 2\mapsto 2, 3\mapsto 1$. Then $1\in f[A]\cap f[B]$, but $1\notin f[A\cap B].$

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Here is a proof:

Let $y\in f[A\cap B]$, that means there is $x\in A\cap B$ such that $y=f(x)$.

This means that $x\in A$ and $x\in B$, furthermore $f(x)\in f(A)$ and $f(x)\in f(B)$.

But this implies that $f(x)$ is in $f(A)$ and in $f(B)$ i.e. $$f(x)\in f(A)\cap f(B) $$

Finally,

$y=f(x)$, therefore $y=f(x)\in f(A)\cap f(B)$

Conclusion:

$$y\in f[A\cap B] \Longrightarrow y\in f(A)\cap f(B) $$

that's the definition of $f[A\cap B] \subseteq f(A)\cap f(B)$

                                                            Q.E.D.

Oops, completely missed the main question :( The equality will happen for $f(x)=x$ if $f:\mathbb{R}\longrightarrow \mathbb{R}$.

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Let $f(x)=x^2$, $A = (-\infty, 0]$, $B=[0, +\infty)$.

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  • $\begingroup$ isn't $f(A)=f(B)=\mathbb{R}^+\cup{\{0\}}$, so $f(A)\cap{f(B)}=\mathbb{R}^+\cup{\{0\}}$ but $f{(A\cap{B})}=f(0)=0$? $\endgroup$ – Iceman Sep 17 '13 at 21:35
  • $\begingroup$ Yes, you are right. $\endgroup$ – njguliyev Sep 17 '13 at 21:37
  • $\begingroup$ but if you just made it the identity mapping it would work right? $\endgroup$ – Iceman Sep 17 '13 at 21:38
  • $\begingroup$ No, for the identity mapping $f[A\cap{B}]=\{0\}$. $\endgroup$ – njguliyev Sep 17 '13 at 21:42
  • $\begingroup$ but $f[A]=A=(-\infty,0]$ and $f[B]=B=[0,\infty)$ if our mapping is $f(x)=x$ so we still get $f[A\cap{B}]=f[A]\cap{f[B]}=0$? $\endgroup$ – Iceman Sep 17 '13 at 21:45

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