# If $f^{-1}(B)$ is compact for all compact $B$, then $f(A)$ is closed for all closed $A$

How to prove this exercise?

Let $f$ be a continuous function of $\mathbb R^n$ to $\mathbb R^m$, so that $f^{-1}(B)$ is compact in $\mathbb R^n$ for all compact $B$ in $\mathbb R^m$. Prove that $f(A)$ is a closed set in $\mathbb R^m$ for all closed set $A$ in $\mathbb R^n$.

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Do not change the question and title completely. Ask a new question. – William Sep 17 '11 at 4:34
@Andres Please do not change the question so substantially that you invalidate the current answers completely. If you want to ask a question about sequences, ask a fresh question; please do not edit the existing question on topology to ask this. – Srivatsan Sep 17 '11 at 4:35

You only need continuity of f; f continuous sends compact sets to compact sets.

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sorry A is a closed set, not a compact set. – Roiner Segura Cubero Sep 17 '11 at 2:31
sorry before I was wrong. I have to prove that if f be a continuos function of R^n to R^m, so that f^{−1}(B) is compact in R^n for all compact B in R^m then f(A) is a closed set in R^m for all closed set A in R^n. Plaese help me with this exercise. – Roiner Segura Cubero Sep 17 '11 at 2:56
Hi! How to prove that all real number x is limit de one sequence of the form {m/2^n} with m a integer number? – Roiner Segura Cubero Sep 17 '11 at 4:04
Andres: the dyadic rationals are a dense subset of the reals. Think of the 4-ple $1/4, 1/2,3/4,1$ and then repeat this process of division for each subinterval, to pin down a number. – MAK Sep 17 '11 at 4:59

The question has since been changed. The next two paragraphs are no longer relevant.

The continuous image of a compact set is compact regardless of your additional property.

The proof is the following. Suppose $\mathcal{C}$ is a covering of $f(A)$ by open sets. Then $f^{-1}(X)$ for any $X \in \mathcal{C}$ is open, since $f$ is continuous. Thus the inverse image of every open set in $\mathcal{A}$ is an open covering of $A$. Since $A$ is compact, a finite number of these cover $A$. Thus a finite number of the original covering cover $f(A)$.

In light of the comments, the question has been phrased incorrect. Given a continuous function with the properties stated in the question, I will prove that $f$ is a closed map. Note I am proving that a proper map between euclidean space is a closed map.

Let $A$ be closed. Let $x$ be a limit point of $f(A)$. Since we are in $\mathbb{R}^m$, we can choose a ball $U$ around $x$ such that its closure $\bar{U}$ is compact. Therefore, $f^{-1}(\bar{U})$ is compact. Compact subsets of euclidean space are closed. Therefore $A \cap f^{-1}(\bar{U})$ is compact since close subsets of compact sets are compact. As mentioned above, continuous image of compact sets are compact. Therefore, $f(A \cap f^{-1}(\bar{U}))$ is compact and hence closed in $\mathbb{R}^m$. Let $Z = f(A \cap f^{-1}(\bar{U}))$. Therefore, $Z$ is closed. $x$ is a limit point of f(A) and $\bar{U}$ contains $x$, so x is a limit point of $Z$. Since, $Z$ is closed, $x \in Z \subset f(A)$. Thus $x \in f(A)$. $f(A)$ contains all its limits point; hence, it is closed.

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sorry A is a closed set, not a compact set. – Roiner Segura Cubero Sep 17 '11 at 2:30
I don't think this is true. Let $n = m$. $\mathbb{R}^n$ is closed. Let $f$ be the identity function The identity function has the desired property. But $f(\mathbb{R}^n) = \mathbb{R}^n$, and this set is not compact. – William Sep 17 '11 at 2:37
sorry I was wrong. I have to prove is that f(A) is a closed set in R^m for all closed set A in R^n. – Roiner Segura Cubero Sep 17 '11 at 2:41
sorry before I was wrong. I have to prove that if f be a continuos function of R^n to R^m, so that f^{−1}(B) is compact in R^n for all compact B in R^m then f(A) is a closed set in R^m for all closed set A in R^n. Plaese help me with this exercise. – Roiner Segura Cubero Sep 17 '11 at 2:56
@Andres: You can edit your question to correct the mistake. – Nate Eldredge Sep 17 '11 at 2:57

Suppose $A$ is closed and that we have a Cauchy sequence $y_n\in f(A)$, it suffices to show that $y_n$ converges to a point of $f(A)$ (in order to prove that $f(A)$ is closed).

Hint:

1. Show that there is a Cauchy sequence $x_n\in A$ such that $f(x_n)=y_n$.

2. What can we say about $f(\lim x_n)$?

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A.D.: I didn't downvote you, but in a lot of introductory topology classes, sequences either aren't used at all or only later on. – Gerben Sep 17 '11 at 12:16
@Gerben: Sure, it is tagged General topology, but deals with the specific spaces $\mathbb{R}$. – AD. Sep 17 '11 at 18:32