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?

  • $\begingroup$ I thought the $\subseteq$ should be $=$ $\endgroup$ – DHMO Apr 16 '17 at 14:06
  • $\begingroup$ @DHMO that's when $f$ is $1-1$. $\endgroup$ – kobe Apr 16 '17 at 14:07
  • $\begingroup$ In general, they are not equal. $\endgroup$ – user370967 Apr 16 '17 at 14:07
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    $\begingroup$ Let $f : \Bbb R \to \Bbb R$ given by $f(x) = x^2$. If $A = [-1,0]$ and $B = [0,1]$, then $f(A\cap B) = f(\{0\}) = \{0\}$, but $f(A)\cap f(B) = [0,1] \cap [0,1] = [0,1] \neq \{0\}$. $\endgroup$ – kobe Apr 16 '17 at 14:10
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    $\begingroup$ @kobe Wow. I need to tune my intuition. $\endgroup$ – DHMO Apr 16 '17 at 14:11

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 $f(A)\cap f(B)=\{0\}$.



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

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 $\cap$.

  • $\begingroup$ I am asking how to logically prove this statement. I know the definitions. $\endgroup$ – Salman Qureshi Apr 16 '17 at 14:07
  • $\begingroup$ @SalmanQureshi Am I clear enough? $\endgroup$ – DHMO Apr 16 '17 at 14:12
  • $\begingroup$ Gimme a few minutes. I am trying to wrap my mind around it. $\endgroup$ – Salman Qureshi Apr 16 '17 at 14:14

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


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 $f(A\cap B) \subseteq f(A) \cap f(B)$.


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 $y \in f(A) \cap f(B)$.


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