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Prove that $f\colon M\to N$ (topological spaces) is closed if and only if for all $y\in N$ and all open sets $V\supset f^{-1}\left(\{y\}\right)$ in $M$ there exists an open set $U$ in $N$ containing $y$ such that $V\supset f^{-1}(U)\supset f^{-1}(\left\{y\right\})$.

I can't prove this in any way. I tried for 3 days.

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    Because $f$ is a closed map, which is defined in terms of what it does to closed sets, and because open sets are used in the alternative description, my inclination is to see what happens if we take complements. E.g. if $f$ is closed and $V$ is an open set containing $f^{-1}\{y\}$, then $N\setminus f(M\setminus V)$ is an open set containing $y$.2012-06-18

4 Answers 4

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I was able to obtain one direction:

Fix some $y\in N$. Note that $V\supseteq f^{-1}(U)\supseteq f^{-1}(\{y\})$ if and only if $V^c\subseteq f^{-1}(U)^c= f^{-1}(U^c)\subseteq f^{-1}(\{y\})^c.$ If $f$ is closed, then for any closed set $C\subseteq M$ disjoint from $f^{-1}(\{y\})$, we have that $D=f(C)$ is a closed set disjoint from $\{y\}$, and $C\subseteq f^{-1}(f(C))=f^{-1}(D)\subseteq f^{-1}(\{y\})^c.$ Then $V=C^c$ and $U=D^c$ are open subsets of $M$ and $N$ respectively satisfying the necessary property, and every open subset $V\supseteq f^{-1}(\{y\})$ of $M$ can of course be obtained as the complement of a closed subset $C$ of $M$ that is disjoint from $f^{-1}(\{y\})$.

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    Thanks very clear, I didn't noted that $f(V^c)$ not contains $y$. I'm not so accustomed to use some properties of functions to solve problems.2012-06-19
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$(\Rightarrow)$ Zev Chonoles has already given this direction. I give the other.

$(\Leftarrow)$ Suppose $C \subset M$ is closed. Want to show $f(C)$ is closed in $N$. Suppose $y$ is a limit point of $f(C)$. First, you want to show that $f^{-1}(\{y\}) \neq \emptyset$. Suppose $f^{-1}(\{y\}) = \emptyset$. Then let $V = \emptyset$. Then for any $U \subset N$ containing $y$, since $y$ is a limit point of $f(C)$, this $U$ must contain a point of $C$. So $f^{-}(U) \neq \emptyset$. Contradiction. It is impossible that $V = \emptyset \supset f^{-1}(U) \supset f^{-1}(\{y\}) = \emptyset$. I have so far shown that $f^{-1}(\{y\}) \neq \emptyset$.

Now I claim is that at least one point in $f^{-1}(y)$ is a limit point of $C$. Suppose not, then every $x \in f^{-1}(y)$ has neighborhood $B_x$ such that $B_x \cap C = \emptyset$. Let $V = \bigcup_{x \in f^{-1}(y)}B_x$. $V$ is a neighborhood of $f^{-1}(y)$ such that $V \cap C = \emptyset$. For every $U \subset N$ containing $y$, $U$ contains a point of $C$ since $y \in U$ is a limit point of $f(C)$. So $f^{-1}(U) \cap C \neq \emptyset$ so it is impossible that $V \supset f^{-1}(U) \supset f^{-1}(\{y\})$. So I have shown that at least one point $z \in f^{-1}(y)$ is a limit point of $C$. Since $C$ is closed, $z \in C$. So $f(z) = y$. So $y \in f(C)$. $f(C)$ contains all its limit points. So it is closed. $f$ is a closed map.

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    I need more time to diggest your answer. I'm not so good working with limit points. I proved $\Leftarrow$ using same idea that Zev Chonoles♦.2012-06-19
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There’s even a pointwise version of the result. I denote the topology of a space $X$ by $\tau(X)$.

For $y\in N$ let

$\begin{align*} \mathscr{N}(y)&=\{U\in\tau(N):y\in U\},\\ \mathscr{B}_y&=\{f^{-1}[U]:y\in U\in\tau(N)\},\text{ and}\\ \mathscr{V}_y&=\{V\in\tau(M):f^{-1}[\{y\}]\subseteq V\}\;, \end{align*}$

and let $\mathscr{F}_y$ be the filter on $M$ generated by $\mathscr{B}_y$; the claim is that $f$ is closed iff for each $y\in N$, $\mathscr{V}_y\subseteq\mathscr{F}_y$.

Definition: The function $f$ is closed at $y\in N$ iff for each closed $K\subseteq M$, $y\in\operatorname{cl}f[K]$ iff $y\in f[K]$.

Proposition: Let $y\in N$; then $f$ is closed at $y$ iff $\mathscr{V}_y\subseteq\mathscr{F}_y$.

Proof: Suppose that $\mathscr{V}_y\nsubseteq\mathscr{F}_y$, and fix $V\in\mathscr{V}_y\setminus\mathscr{F}_y$. Let $K=M\setminus V$; $K$ is closed, and $K\cap f^{-1}[\{y\}]=\varnothing$, so $y\notin f[K]$. Since $V\notin\mathscr{F}_y$, for each $U\in\mathscr{N}(y)$ there is a point $x_U\in f^{-1}[U]\setminus V=K\cap f^{-1}[U]$; clearly $f(x_U)\in U\cap f[K]$, so $y\in\operatorname{cl}f[K]$, and therefore $f[K]$ is not closed.

Conversely, suppose that $y\in\operatorname{cl}f[K]\setminus f[K]$ for some closed $K\subseteq M$. Let $V=M\setminus K$; clearly $V\in\mathscr{V}_y$. Suppose that $U\in\mathscr{N}(y)$; $y\in\operatorname{cl}f[K]$, so $U\cap f[K]\ne\varnothing$, and therefore $f^{-1}[U]\setminus V=f^{-1}[U]\cap K\ne\varnothing\;.$ Thus, $V\notin\mathscr{F}_y$, and hence $\mathscr{V}_y\nsubseteq\mathscr{F}_y$. $\dashv$

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    I need more time to diggest your answer. What is the filter generated by a set?2012-06-19
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I think that I have the other direction using the same idea that Zev Chonoles ♦:

Let $F\subset M$ a closed subset. Let some $y\in f[F]^c$ then $f^{-1}[\{y\}]\subset f^{-1}[f[F]^c]=f^{-1}[f[F]]^c\subset F^c,$ in particular $f^{-1}[\{y\}]\subset F^c$ where $F^c$ is open. Therefore exists an open $U$ containing $y$ such that $f^{-1}[U]\subset F^c\implies F\subset f^{-1}[U^c]$ therefore $f[F]\subset f[f^{-1}[U^c]]\subset U^c,$ in particular $y\in U\subset f[F]^c.$ Therefore $f[F]^c$ is open and $f[F]$ is closed.