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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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    @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**.2011-09-17

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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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    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?2011-09-17
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You only need continuity of f; f continuous sends compact sets to compact sets.

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    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.2011-09-17
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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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    @Gerben: Sure, it is tagged General topology, but deals with the specific spaces $\mathbb{R}$.2011-09-17