Proof of $f(A∩B)⊆f(A)∩f(B)$ Let there be a function $f:R→R$. Let $A$ & $B$  be two subsets of $R$. From my experience, I know that $f(A∩B)⊆f(A)∩f(B)$.
I don't know how to prove it. Also, are there certain conditions that the function $f$ or the sets $A$ & $B$ should follow for the above condition to be true?
 A: Take $y\in f(A\cap B)$. Then, by definition, there exists $x\in A\cap B$ such that $y=f(x)$;


*

*since $x\in A$, we have $y=f(x)\in f(A)$,

*since $x\in B$, we have $y=f(x)\in f(B)$.


Therefore $y=f(x)\in f(A)\cap f(B)$
Equality does not necessarily hold. Take $f\colon\{1,2\}\to\{0\}$ (the only possible map); take $A=\{1\}$ and $B=\{2\}$. Then $f(A\cap B)=f(\emptyset)=\emptyset$, whereas $f(A)\cap f(B)=\{0\}$.
A: Definitions:


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*$f(A \cap B) = \{f(x): x \in A \cap B\}$

*$f(A) \cap f(B) = \{f(x): x \in A\} \cap \{f(x): x \in B\}$


Now, $A \cap B \subseteq A$ and $A \cap B \subseteq B$.
Therefore, $f(A \cap B) \subseteq f(A)$ and $f(A \cap B) \subseteq f(B)$.
And the result follows from messing around with the logical definitions of $\subseteq$ and $\cap$.
A: A proof by elements is the safe way:
Let $y\in f(A\cap B)$. By definition, $y = f(x)$ for some $x\in A\cap B$. Therefore $f(x)\in A$ and $f(x)\in B$, which means $y=f(x)\in f(A)\cap f(B)$.
A: Let $y \in f(A \cap B)$. Then $y=f(x)$ for some $x \in A \cap B$.
Since $x \in A$ and $y=f(x)$ we get $y \in f(A)$.
Since $x \in B$ and $y=f(x)$ we get $y \in f(B)$.
Therefore, $y \in f(A) \cap f(B)$.
This shows $f(A\cap B) \subseteq f(A) \cap f(B)$.
A: Directly by definition you can prove it.
Let $y \in f(A \cap B)$. (This is because $X \subset Y$ means every element of $X$ is of $Y$; so to prove $X \subset Y$ we take an arbitrary one out of $X$ to see if that one is in $Y$.) Then by definition there is some $x \in A \cap B$ ensuring $y = f(x)$. Then note that $x \in A \cap B$ implies $x \in A$ and $x \in B$, implying that $f(x) \in f(A)$ and $\in f(B)$. So $y \in f(A) \cap f(B)$.
