Proper function sends closed sets in closed sets

Question

Let $X$, $Y$ be metric spaces and let $f:X \rightarrow Y$ be a continuous function. We say that $f$ is a proper function if inverse images of compact sets in $Y$ are compact sets in $X$. I need proof that, if $f$ is a proper function, then $f$ sends closed sets of $X$ in closed sets of $Y$.

Well, my idea was the following: Let $C \subset X$ be a closed set and let $\{y_n \}_{n \in \mathbb{N}}$ be a convergent sequence such that $y_n \in f(C)$ for every $n \in \mathbb{N}$. So, I have to proof that $\lim_{n} y_n = y \in f(C)$.

Note that $K = A \cup \{y \}$, where $A = \{y_n : n \in \mathbb{N} \} \subset f(C)$, is a compact set in $Y$ (besides that, A is compact too). As $f$ is proper, $f^{-1}(K)$ and $f^{-1}(A)$ are compacts in $X$.

The problem is that I don't know how to go on. What could I say about the 'sequence' $f^{-1}(A) = \bigcup_{n \in \mathbb{N} }f^{-1}(y_n)$ in $C \subset X$ ?

Can someone help me?

Have someone another idea to the proof?

Is this result valid to topological spaces??

The result is not true without some assumptions on the spaces $X$ and $Y$. For example map a finite set with the discrete topology onto a another finite set with the trivial topology. — Rene Schipperus, Oct 07 '16 at 02:53
For counterexamples in non-Hausdorff spaces see this question and this one. — Brian M. Scott, Oct 07 '16 at 08:37

score 2 · Accepted Answer · answered Oct 07 '16 at 02:24

2

Let $C$ be a closed subset and $y$ an element in the adherence of $f(C)$, there exists $y_n=f(x_n), x_n\in C$ such that $lim_ny_n=y$, $A=\{y_n, n\in N\}\bigcup\{y\}$ is compact. This implies that $f^{-1}(A)$ is compact. This implies that there exists a subsequence $x_{n_i}$ of $x_n$ which converges. Since $C$ is closed, $x\in C$. Since $f$ is continue, $f(x)=limf(x_{n_i})=y$. We deduce that $y\in f(C)$ and $f(C)$ is closed.

answered Oct 07 '16 at 02:24

Tsemo Aristide

89,587

You are assuming that the spaces are metric or something similar. It is not true that a an element of the adherence is a limit of a sequence nor is it true in general that a sequence in a compact set has a convergent subsequence. A counter example is ${0,1}^{[0,1]}$. – Rene Schipperus Oct 07 '16 at 02:56
The author of the question in his proof attempt assume that too,... I believe that the spaces are separated. My proof is just a continuation of his idea. – Tsemo Aristide Oct 07 '16 at 02:58
Sorry guys. In fact, $X$ and $Y$ are metric spaces. I thought it could be generalized, but I made a mistake. – Rodrigo Dias Oct 07 '16 at 03:24

Proper function sends closed sets in closed sets

1 Answers1