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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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    I think this (freely available) paper may be useful: http://www.ams.org/journals/proc/1970-024-04/S0002-9939-1970-0254818-X/S0002-9939-1970-0254818-X.pdf Anyway, the solutions below look correct.2012-04-26

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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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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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    @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 $p$.2012-04-26
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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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    To get a contradiction, I am assuming that for all \epsilon > 0, the set $B(\hat y,\epsilon)$ intersects with the set $f(\mathbb{R}^n)$. Choose $\epsilon = \frac{1}{k}$, pick any point in the intersection and call it $y_k$. It should be clear that the sequence converges to $\hat y$.2012-04-26
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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 $f(\mathbb{R}^n) = \mathbb{R}^m$.