First let me say I am aware of the other threads on this result. The reason for me making this thread is to find out whether or not my proof/proof attempt is correct.

The problem stated in full detail is given below.

Let $X,Y$ be sets and $f:X \rightarrow Y$, let $C,D \subseteq X$ and let $A,B \subseteq Y$. Prove that $f^{-1}(A) \cap f^{-1}(B)= f^{-1}(A \cap B)$.

Here is my attempt:

$f^{-1}(A) \cap f^{-1}(B)= { \{x\in X \mid f(x) \in A\}}\cap { \{x\in X \mid f(x) \in B\}}= \{x\in X \mid f(x) \in A\wedge f(x) \in B\}= \{x\in X \mid f(x) \in A\cap B \}=f^{-1}(A \cap B)$

If anyone would be kind enough to explain to me where I've gone wrong or made an unjustified assumption I would be very grateful!

P.S. Sorry about the bad formating

  • $\begingroup$ you have only proved $f^{-1}(A) \cap f^{-1}(B)\subseteq f^{-1}(A \cap B)$. You will need to prove that $f^{-1}(A \cap B)\subseteq f^{-1}(A) \cap f^{-1}(B)$ $\endgroup$ – Neo Oct 16 '18 at 11:32
  • $\begingroup$ @Neo Really? As far as I can tell I have only used equality ad not implication rules... $\endgroup$ – krtgdl Oct 16 '18 at 11:33
  • $\begingroup$ @Neo Nope. Every equality OP wrote is true, they proved equality, not just inclusion. $\endgroup$ – 5xum Oct 16 '18 at 11:35
  • $\begingroup$ Learnt something new, I was marked wrong in an assignment for doing that once :/ $\endgroup$ – Neo Oct 16 '18 at 11:37
  • $\begingroup$ @Neo If you did it the wrong way, it might have been marked correctly. It's possible to use the correct tool incorrectly. Or maybe you were cheated. I'd have to see the concrete example. $\endgroup$ – 5xum Oct 16 '18 at 11:56

Your proof seems perfectly OK to me.

If you aren't sure because the set notation is confusing you, maybe rewriting it without so many sets would also work:

$$\begin{align}x\in f^{-1}(A)\cap f^{-1}(B)\iff &(x\in f^{-1}(A)\land x\in f^{-1}(B)\\\iff& f(x)\in A\land f(x)\in B\\\iff& f(x)\in A\cap B\\\iff& x\in f^{-1}(A\cap B)\end{align}$$


Your proof is fine ! Everything is correct.


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