Please who can explain why $$f(A \cup B) = f(A) \cup f(B)$$ (And I know how to prove it using set theory symbols ) But $$f(A \cap B) \subseteq f(A) \cap f(B)$$

And the equality arises if and only if $f$ is injective.

Thanks in advance

  • 4
    $\begingroup$ Use element chasing. Suppose $y\in f(A\cup B)$. What does that mean by the definition of what $f(A\cup B)$ is in the first place? It means that there must be some $x\in A\cup B$ such that $f(x)=y$. Now... since $x\in A\cup B$ there are two (not exclusive) possibilities, we either have $x\in A$ or we have $x\in B$. In the first case it follows that $y\in f(A)$ and so $y\in f(A)\cup f(B)$. In the second case, similar things occur. That shows $f(A\cup B)\subseteq f(A)\cup f(B)$. Now, do the same thing but in reverse to show the opposite inclusion and therefore equality. $\endgroup$
    – JMoravitz
    Jan 1, 2018 at 16:40
  • $\begingroup$ OK thanks I'll do just that $\endgroup$ Jan 1, 2018 at 18:36
  • 1
    $\begingroup$ Example. Suppose $a\ne b$ and $f(a)=f(b)$. Let $A=\{a\}$ and $B=\{b\}.$ $\endgroup$ Jan 1, 2018 at 18:38

3 Answers 3


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

$x\in A\cap B \implies x\in A$ and $x\in B \implies f(x) \in$ $f(A)\cap f(B)$.

converse need not be true. Consider $f(x)=\sin x, A=[0,\frac{\pi}{2}]$, $B=[\frac{\pi}{2},\pi]$, $f(B)=f(A)=[0,1]$.

Suppose $f$ is injective, $y \in f(A)\cap f(B)$. That is $y=f(a), a \in A$ and $y=f(b), b \in B$. $\because f$ is injective $\implies$ $a=b$ $\implies$ $y\in f(A\cap B).$

  • $\begingroup$ @JoseCarlosSantos Thank you sir $\endgroup$
    – user464147
    Jan 1, 2018 at 18:12

First we show $$f(A \cup B) = f(A) \cup f(B)$$ which is easier to explain.If $$y\in f(A\cup B)$$ then $y=f(x)$ for some $x\in A\cup B.$ That is $x\in A $ or $x\in B$.Therefore $f(x)\in f(A)$ or $f(x)\in f(b)$ that is $$y=f(x)\in f(A)\cup f(B).$$ On the other hand if $$y\in f(A)\cup f(B).$$then $y\in f(A)$ or $y\in f(B)$. If $y\in f(A)$, there exists an $x\in A$ such that $f(x)=y$ and if $y\in f(B)$then there exists an $x\in B$ such that $ f(x)=y$. In either case $x\in A\cup B$ which implies $$y=f(x)\in f(A\cup B).$$ Now we get to the $$f(A \cap B) \subseteq f(A) \cap f(B)$$If $$y\in f(A\cap B)$$then $y=f(x)$ for some $x\in (A\cap B).$ That is $x\in A $ and $x\in B$.Therefore$f(x)\in f(A)$ and $f(x)\in f(b)$ that is $$y=f(x)\in f(A)\cap f(B).$$ Now we get to the tricky part. If $$y=f(x)\in f(A)\cap f(B).$$ then $y\in f(A)$ and $y\in f(B)$ That is for some $x_1$ in $A$ we have $ f(x_1)=y$ and for some $x_2\in B$ we have $ f(x_2)=y.$ Since $x_1=x_2$ if and only if $f$ is injective, we can not include that $y\in f(A\cap B)$ unless $f$ is injective.


An element of $f(A) \cap f(B)$ can come from a single element in $A \cap B$ or two distinct elements in $A \cap B$. An element of $f(A \cap B)$ can only come from one distinct element in $A \cap B$.

This can be shown using the logic of set membership and intersections.

For a function $f:X \to Y$, the set membership logic for an element $y \in Y$ is:

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space y \in f(X) \iff y = f(x)$ for some $x \in X $

From here on, "for some $x \in X$" (or $a \in A$, etc), will be implied when $f(x)$ (or $f(a)$, etc) appears, to make reading easier. The logic of intersections is: $$x \in A \cap B \iff (x \in A) \wedge (x\in B) $$

Then for a function $f:(A \cup B) \to X$, the intersection of the images is: $$ x \in f(A) \cap f(B) \iff (x \in f(A)) \wedge (x \in f(B)) $$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\iff (x = f(a)) \wedge (x = f(b))$

Note that this carries the implications $f(a) = f(b) = x$ and $a,b \in A \cap B$, but does not require that $a = b$. So any pair of points $(a,b)$ in $A \cap B$ such that $f(a) = f(b)$, distinct or not, corresponds to an element $x$ in $f(A) \cap f(B)$. But if we take the image of the intersection, we get:

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space x \in f(A \cap B) \iff x = f(c)$ for some $c \in A \cap B$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\iff (x = f(a)) \wedge (x=f(b)) \wedge (a=b)$

$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\iff (x \in f(A) \cap f(B)) \wedge (a=b)$

... And we see that the image of the intersection is the intersection of the images with the restriction that only single points in $A \cap B$ (as opposed to pairs of points) correspond to the members of $f(A) \cap f(B)$. Therefore, $x \in f(A \cap B) \to x \in f(A) \cap f(B)$, which in terms of sets defines $f(A \cap B) \subseteq f(A) \cap f(B)$.

However if $f$ is an injection, then $f(a) = f(b) \to a = b$, and then $x \in f(A) \cap f(B) \iff x \in f(A \cap B)$, which in terms of sets defines the equality $f(A) \cap f(B) = f(A \cap B)$.

To show that $f(A) \cap f(B) = f(A \cap B)$ implies $f$ is an injection, we noted that in logical terms this equality means that for all $x \in X$: $$x \in f(A) \cap f(B) \iff x \in f(A \cap B)$$ $\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(x = f(a)) \wedge (x=f(b)) \iff (x = f(a)) \wedge (x=f(b)) \wedge (a=b)$

Again, since both sides imply $a,b \in A \cap B$, we have for all $a$ and $b$ in $A \cap B$: $$(f(a) = f(b)) \iff (f(a) = f(b)) \wedge (a = b)$$

which means $(f(a) = f(b)) \to (a = b)$ (you can check that $(P \iff P \wedge Q) \to (P \to Q)$ is a tautology). This is the contrapositive equivalent of the condition for an injection, which is $a \neq b \to f(a) \neq f(b)$


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