Ok. I have managed thru:
- If f is continuous then the preimage of open set is a open set
- If the preimage of the open set is a open set then f is continuous
- If f is continuous then the preimage of a closed set is closed.
Silly me, I'm struggling with:
- If the preimage of a closed set is closed then f is continuous.
The proof I'm working with is:
Let $ f: X \to Y $ and let $ U \subset Y $ a open set therefore we have $f^{-1}(U) $ a open set (by statement 1)
Lets take $Y \setminus U$, this set is closed (Ok here)
Therefore $ f^{-1}(Y\setminus U) $ is closed, by hypothesis (Ok here)
Now the part that I'm struggling. (not funny)
$ f^{-1}(Y \setminus U) = X \setminus f^{-1}(U) $
If I accept this affirmation, ok, the proof finishes because I can tatke the complement . But I can't stand this affirmation can be true from nowhere. Let me explain my reasoning.
I now that I can map $ U $ open, to a $f^{-1}(U)$ an open set (already proved). But if I take the complement of $U$ I will take everything else from he image, how can I garantee that it will map to everything else from the domain (besides the $f^{-1}{U}$), I didn't said anything if the mapping is onto or bijective. Therefore I think that $f^{-1}(Y \setminus U) \subset X$ but I can't garantee $f^{-1}(Y \setminus U) \cup f^{-1}(U) = X $ and
$f^{-1}(Y \setminus U) \cap f^{-1}(U) = \emptyset $.
Can someone enlighten me?
Best Answer
By definition $$f^{-1}(A):=\{x\in X\mid f(x)\in A\}$$In general: $$x\in f^{-1}(A)\iff f(x)\in A$$
Statement $x\in f^{-1}(Y\setminus U)\vee x\in f^{-1}(U)$ is equivalent with statement $f(x)\in Y\setminus U\vee f(x)\in U$, wich is evidently true for every $x\in X$. So: $$f^{-1}(Y\setminus U)\cup f^{-1}(U)=X$$
Statement $x\in f^{-1}(Y\setminus U)\wedge x\in f^{-1}(U)$ is equivalent with statement $f(x)\in Y\setminus U\wedge f(x)\in U$, which is evidently not true for any $x\in X$. So: $$f^{-1}(Y\setminus U)\cap f^{-1}(U)=\varnothing$$