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I am a new user in Math Stack Exchange. I don't know how to solve part of this problem, so I hope that one of the users can give me a hand.

Let $f$ be a continuous function from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$ with the following properties:$A\subset \mathbb{R}^{n}$ is open then $f(A)$ is open. If $B\subset \mathbb{R}^{m}$ is compact then $f^{-1}(B)$ is compact.

I want to prove that $f( \mathbb{R}^{n}) $ is closed.

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Here's an idea: let $B$ be compact. Then by Heine-Borel you know that it's closed and bounded. Hence you know $f^{-1}B$ is closed. It remains to be shown that $f^{-1}B$ is bounded. By contradiction assume that it's not. Then take a sequence in it and show that $B$ would not be bounded either. –  Matt N. Apr 26 '12 at 17:11
    
@MattN.: He's not trying to prove that the inverse image of a compact set is compact; this is assumed. –  Arturo Magidin Apr 26 '12 at 17:13
    
@ArturoMagidin Doh. Too quick on the trigger. Thanks. I'll leave the comment anyway. –  Matt N. Apr 26 '12 at 17:14
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I think this (freely available) paper may be useful: ams.org/journals/proc/1970-024-04/S0002-9939-1970-0254818-X/… Anyway, the solutions below look correct. –  Siminore Apr 26 '12 at 17:34

4 Answers 4

up vote 1 down vote accepted

Try another approach; show that the complement of $f(\mathbb{R}^n)$ is open.

This is trivial if the complement is empty, so suppose the complement is not empty, and choose $\hat y \notin f(\mathbb{R}^n)$. You want to show that there exists some $\epsilon>0$ such that the set $B(\hat y, \epsilon)$ also lies in the complement.

You can proceed by contradiction and generate a sequence of points $y_k \in f(\mathbb{R}^n)$ that converge to $\hat y$.

Now consider the set $\{y_k\} \cup \{\hat y\}$. What properties does it have in relation to the second property above, and how does this lead to a contradiction?

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I am not sure if I understand your proof, but here is what I could see in your proof: The set $\{y_k\} \cup \{\hat y\}$ is a set of points in $\mathbb{R}^{m}$ so it is closed and bounded and hence it is compact, and by the second property the preimage of this set in $\mathbb{R}^{n}$ is compact as well. I can't see where the contradiction come from? –  mchris619 Apr 26 '12 at 17:36
    
Also, how can you be sure that such a sequence $y_{k}$ exists? –  mchris619 Apr 26 '12 at 17:38
    
(1) Note that since $y_k \in f(\mathbb{R}^n)$, then there must be an $x_k$ such that $y_k = f(x_k)$. Since the $x_k$ lie in $f^{-1}(\{y_k\} \cup \{\hat y\})$ which is compact, then $x_k$ must have a convergent subsequence. What can you say about the limit since $f$ is continuous? (2) The contradiction works by assuming that $B(\hat y,\epsilon)$ intersects $f(\mathbb{R}^n)$ for all $\epsilon >0$. If this is not true, then the complement is open. –  copper.hat Apr 26 '12 at 17:47
    
By continuity of $f$, I conclude that $f(x_{k})\rightarrow $ converges to ${\hat y\}$. –  mchris619 Apr 26 '12 at 17:59
    
Correct, which would contradict the choice of $\hat y$ in the first place. So we conclude that there must exist some $\epsilon >0$ such that $B(\hat y, \epsilon)$ is contained in the complement. Hence the complement is open, which means that $f(\mathbb{R}^n)$ is closed. –  copper.hat Apr 26 '12 at 18:01

I’ve left some of the details to you, but here’s the main outline.

Suppose that $f[\Bbb R^n]$ is not closed in $\Bbb R^m$. Then there is a point $p\in\operatorname{cl}f[\Bbb R^n]\setminus f[\Bbb R^n]$, and there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $f[\Bbb R^n]$ converging to $p$. Let $K=\{p\}\cup\{x_n:n\in\Bbb N\}$, and show first that $K$ is compact, so that $f^{-1}[K]$ is compact in $\Bbb R^n$. Now for each $n\in\Bbb N$ choose $y_n\in f^{-1}[K]$ such that $f(y_n)=x_n$, and consider the sequence $\langle y_n:n\in\Bbb N\rangle$. This is a sequence in the compact set $f^{-1}[K]$, so it has a convergent subsequence $\langle y_{n_k}:k\in\Bbb N\rangle$, say with limit $y$. What must $f(y)$ be? Why is this a contradiction?

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$f(y)$ should be $p$? but I really can't see where the contradiction is? –  mchris619 Apr 26 '12 at 17:50
    
@mchris619: Yes, $f(y)$ must be $p$, because $f$ is continuous and $\langle x_{n_k}:k\in\Bbb N\rangle\to p$. But where did we choose $p$? Can $p$ be $f(y)$? –  Brian M. Scott Apr 26 '12 at 17:58
    
You're right, $p$ cannot be $f(y)$ because $p$ is chosen from the beginning to be in the complement of $f( \mathbb{R}^{n}) $. But how can you justify the existence of the sequence ${x_{n}}$ that converges to $p$. By whcih theorem does this follow? –  mchris619 Apr 26 '12 at 18:20
    
@mchris619: Recall that $p\in\operatorname{cl}f[\Bbb R^n]\setminus f[\Bbb R^n]$. Since $p\in\operatorname{cl}f[\Bbb R^n]$, for each $n\in\Bbb N$ there must be some $x_n\in f[\Bbb R^n]$ such that $\|x_n-p\|<2^{-n}$. Clearly $\langle x_n:n\in\Bbb N\rangle$ must converge to $p$. –  Brian M. Scott Apr 26 '12 at 18:29

Take $y \in \overline{f(\mathbb{R}^n)}$. Let $B_\varepsilon = \{x | d(x,y) \leq \varepsilon\}$. Now, $\emptyset \neq B_\varepsilon \cap f(\mathbb{R}^n) = f\left(f^{-1}(B_\varepsilon)\right)$. Because $f^{-1}(B_\varepsilon)$ is compact, $B_\varepsilon \cap f(\mathbb{R}^n)$, as the image of a compact by $f$, is a decreasing sequence of nonempty compact sets. Therefore, $\bigcap_\varepsilon (B_\varepsilon \cap f(\mathbb{R}^n))$ is nonempty. Now, $\emptyset \neq \bigcap_\varepsilon (B_\varepsilon \cap f(\mathbb{R}^n)) \subset \bigcap_\varepsilon B_\varepsilon = \{y\}$ implies that $y \in f(\mathbb{R}^n)$.

That is, $f(\mathbb{R}^n) = \overline{f(\mathbb{R}^n)}$.

By the way, the only "clopen" sets in $\mathbb{R}^m$ are $\emptyset$ and $\mathbb{R}^m$. Since $f(\mathbb{R}^n)$ is not empty, we have that $f(\mathbb{R}^n) = \mathbb{R}^m$.

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I think that the most instructive proof is that of Proposition 5.3 of this file. Indeed, it shows that proper maps between locally compact topological spaces are always closed. This is a generalization of this discussion. I find it interesting since "standard" proofs tend to use sequences, and one might believe that everything might be lost without a metric.

By the way, a very nice corollary of the proposed exercise is that non-constant maps with the two properties are always surjective, since $f(\mathbb{R}^n)$ is connected.

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