Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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. – Rudy the Reindeer 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. – Rudy the Reindeer Apr 26 '12 at 17:14
I think this (freely available) paper may be useful:… Anyway, the solutions below look correct. – Siminore Apr 26 '12 at 17:34
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?

share|cite|improve this answer
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?

share|cite|improve this answer
$f(y)$ 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 $f(\mathbb{R}^n) = \mathbb{R}^m$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.