how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

Question

I am given this equation: $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

I want to prove it: what i did is

I take any $a \in f^{-1}(B_1 \cap B_2)$, then there is $b \in (B_1 \cap B_2)$ so that $f(a)=b$. Because of $b \in (B_1 \cap B_2)$, it is true that $b \in B_1$ and $b \in B_2$, so $a \in f^{-1}(B_1)$ and $a \in f^{-1}(B_2)$.

this means $f^{-1}(B_1 \cap B_2) \subseteq f^{-1}(B_1) \cap f^{-1}(B_2)$.

is it ok?

score 11 · Answer 1 · answered Oct 06 '13 at 18:11

Here is essentially the first answer, written out a bit more formally.

Let's start with the basic property of $\;f^{-1}[\cdot]\;$: $$ a \in f^{-1}[B] \;\equiv\; f(a) \in B $$ for any $\;a,B\;$. Using this, we can simply calculate \begin{align} & a \in f^{-1}[B_1 \cap B_2] \\ \equiv & \;\;\;\;\;\text{"the above basic property"} \\ & f(a) \in B_1 \cap B_2 \\ \equiv & \;\;\;\;\;\text{"definition of $\;\cap\;$"} \\ & f(a) \in B_1 \;\land\; f(a) \in B_2 \\ \equiv & \;\;\;\;\;\text{"the above basic property, twice"} \\ & a \in f^{-1}[B_1] \;\land\; a \in f^{-1}[B_2] \\ \equiv & \;\;\;\;\;\text{"reintroduce $\;\cap\;$ using its definition"} \\ & a \in f^{-1}[B_1] \cap f^{-1}[B_2] \\ \end{align} By set extensionality, the statement in question follows.

...and this would even be simpler if the proof was done in the other direction, starting with the most complex side of the equation. — MarnixKlooster ReinstateMonica, Jun 11 '22 at 12:24

score 10 · Accepted Answer · answered Oct 06 '13 at 08:14

10

Yeah...this can be actually written in this way;

$a\in f^{-1}(B_1\cap B_2)$, means $f(a)\in B_1\cap B_2$ and so $f(a)\in B_1$ and $f(a)\in B_2$. Hence, $a\in f^{-1}(B_1)$ and $a\in f^{-1}(B_2)$

answered Oct 06 '13 at 08:14

Anupam

4,908

this also means that $B_1=B_2$ is either the same or it's interesection is empty – Rainb May 07 '23 at 10:40

how to prove $f^{-1}(B_1 \cap B_2) = f^{-1}(B_1) \cap f^{-1}(B_2)$

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