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Let $f:A \to B$ and $C,D\subset A$ then we have the following properties:

$ \space i)\space f(C\cup D)=f(C)\cup f(D)$

$ii)\space f(C\cap D)=f(C)\cap f(D)$

I tried to prove the first property with a direct proof, and the second property with an element-wise proof.

Proof of $i)$

$f(C\cup D)=\{f(x)\in B:x\in C\cup D\}$ (by the definition of a functions's image)

$\space \space \space \space \space \space\space \space \space \space \space \space \space \space \space \space =\{f(x)\in B:x\in C \text{ or } x\in D\}$ ( just writing the set's property in words )

$ \space \space \space \space \space \space\space \space \space \space \space \space \space \space \space \space =\{f(x)\in B:x\in C \}\cup \{f(x)\in B:x\in D\}$ (by the definition of a union )

$\space \space \space \space \space \space\space \space \space \space \space \space \space \space \space \space = f(C)\cup f(D)$ (again by the definition of a functions image)

Proof of $ii)$

Let $f(x)\in f(C\cap D)$, then $x\in C\cap D$ hence $x$ maps to both $f(C)$ and $f(D)$, so $f(x)\in f(C)\cap f(D)$. Conversely, let $f(x)\in f(C)\cap f(D)$ then $x\in C$ and $x\in D$, and therefore $x$ maps to $f(C\cap D)$, hence $f(x)\in f(C \cap D)$. Since we have shown that $f(C \cap D) \subset f(C)\cap f(D)$ and that $f(C)\cap f(D) \subset f(C \cap D) $, we have shown that $ f(C\cap D)=f(C)\cap f(D)$.

I am not sure if the second proof is sufficient. I feel more confident with the first proof. If you could please give my some constructive feedback. Thank you for your time.

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The second proof is wrong:

Take $C=\{c\}$ and $D=\{d\}$, $c\neq d$ and suppose that $f(c)=f(d)=y$ so we have $$f(C\cap D)=\emptyset\neq \{y\}=f(C)\cap f(D)$$

The mistake in your proof is: "Conversely, let $f(x)\in f(C)\cap f(D)$ then..." and you should correct this by: let $y\in f(C)\cap f(D)$ then there's $c\in C$ and $d\in D$ such that $f(c)=f(d)=y$ so in other word the preimage of $y$ is not necessary unique.

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  • $\begingroup$ I see the light! Thanks I see the clearly what I did wrong. $\endgroup$ – JimmyJackson Jul 4 '13 at 9:04
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jul 4 '13 at 9:40
  • $\begingroup$ Nice catch (of the mistake)! $\endgroup$ – Namaste May 26 '14 at 15:43
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This is very elementary but I believe it works for 2. I f we assume $$X=\{a,b\},~~Y=\{c\}$$ and $A_1=\{a\},~~A_2=\{b\}$. If we define $f:X\to Y$ to be constant such that $f(a)=f(b)=c$, so you can easily seen that $f(A_1\cap A_2)=f(\emptyset)=\emptyset$ but $$f(A_1)\cap f(A_2)=\{c\}$$

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Your first proof looks great. Your second proof is missing something though (because, in fact, the claim isn't true), since the $x$'s might not match. That is, we only know that: $$ y \in f(C) \text{ and } y\in f(D) \implies \exists x_1\in C \text{ such that } f(x_1)=y \text{ and } \exists x_2\in D \text{ such that } f(x_2)=y $$ In general, $f(C \cap D) \subseteq f(C) \cap f(D)$. To get equality, you must additionally require that $f$ be injective. This allows you to infer that: $$ f(x_1)=y=f(x_2) \implies x_1=x_2 \implies x_1 \in C \cap D \implies y=f(x_1) \in f(C \cap D) $$

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  • $\begingroup$ So, without assuming that $f$ is injective we cannot prove property two right? $\endgroup$ – JimmyJackson Jul 4 '13 at 8:59
  • $\begingroup$ @JimmyJackson Yes, that's correct. $\endgroup$ – Adriano Jul 4 '13 at 15:55

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