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Let $f:X\to Y$ be a continuous map between metric spaces satisfying the Heine-Borel theorem. Show that $f$ is proper if the following condition holds:

For every sequence $x_n\in X$ such that $f(x_n)$ is bounded, $x_n$ is also bounded.

My definition of properness is that $f^{-1}(K)$ is compact for all compact set $K\subseteq Y$.

Try: Let $K\subseteq Y$ be compact. Then $f^{-1}(K)$ is compact if every sequence in $f^{-1}(K)$ has a convergent subsequence. Let $x_n\in f^{-1}(K)$. Then, $f(x_n)\in K$ which is compact, so there is a convergence subsequence of $f(x_n)$, i.e. $f(x_{n_k})$ converges for some $n_k$. Then, is $x_{n_k}$ bounded?

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Raubeurght Hagh
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  • Were you asked to prove this, or to prove or disprove it? Hint: If $d$ is a metric on $X$ then there exists a metric $d'$ on $X$ which generates the same topology as $d$, and is such that $0\le d(x,y)\le 1$ for all $x,y\in X$. – David C. Ullrich Apr 30 '16 at 12:59
  • @DavidC.Ullrich maybe he uses the notion "bounded = contained in a compact set", which some texts use? Then the sequence $(f(x_n))$ is by the definition bounded. – Henno Brandsma Apr 30 '16 at 13:33
  • I think I am supposed to use the Heine-Borel theorem. – Raubeurght Hagh Apr 30 '16 at 14:02
  • Such spaces are called Borel compact and such spaces are exactly such that bounded sequences have convergent subsequences. – Henno Brandsma Apr 30 '16 at 14:12
  • @DavidC.Ullrich I think the point is that this is a question about metric spaces, not metrizable spaces. – Raubeurght Hagh Apr 30 '16 at 14:26
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    @RaubeurghtHagh Now that you've added the condition that the space have the Heine-Borel property it's certainly a question about metric spaces. Also it's now true. In the original version there was no mention of H-B - if the original version were true (and if "bounded" meant bounded in the metric, as I assumed) it would show that every map from one metric space to another was proper, precisely because any metric is topolgically equivalent to a bounded metric (and the definition of "proper" is a topological property). – David C. Ullrich Apr 30 '16 at 14:34
  • Thank you very much for the clarifications @DavidC.Ullrich. – Raubeurght Hagh Apr 30 '16 at 14:55
  • @DavidC.Ullrich Did you downvoted? – Raubeurght Hagh Apr 30 '16 at 14:56

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Assuming "bounded" means totally bounded, i.e. contained in a compact set, or that $X$ is Borel compact, which also implies that bounded sets are totally bounded:

$K \subseteq Y$ compact, pick a sequence $(x_n)$ in $f^{-1}[K]$. Then $f(x_n)$ has a convergent subsequence in $K$, so there is some $y \in K$ and a subsequence $(f(x_{n_k}))$ that converges to $y$.

Then the $f(x_{n_k})$ sequence is bounded (using my interpreted definition of boundedness) so the $x_{n_k}$ are a bounded sequence so there is some convergent subsequence $(x_{n_{k_l}})$ in it that converges to some $x$. But by continuity of $f$, we know that $f(x_{n_{k_l}})$ converges to $f(x)$, but also to $y$ (as a subsequence of the sequence $(f(x_{n_k}))$. By unicity of limits (in metric spaces certainly) we have $f(x) = y$, so $x \in f^{-1}[K]$. This shows that the original sequence from $f^{-1}[K]$ has a convergent subsequence in $f^{-1}[K]$, so $f^{-1}[K]$ is compact. So $f$ is proper.

The reverse: suppose $f$ is proper. Let $(x_n)$ be a sequence such that $f(x_n)$ is bounded (hence totally bounded, so contained in a compact set $K$ (its closure will do)). Then $x_n$ all lie in $f^{-1}[K]$ which is compact, so $(x_n)$ is bounded.

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Henno Brandsma
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  • Thanks! For example, are manifolds Borel compact? – Raubeurght Hagh Apr 30 '16 at 14:13
  • @RaubeurghtHagh It depends. If they're embeddable as closed sets in Euclidean spaces, they are, I think, giving them the inherited metric. But there are non-metrisable manifolds, but not all people consider these "manifolds". So your definition matters here. – Henno Brandsma Apr 30 '16 at 14:16
  • If I understand correctly: the definition of Borel compact depends not just on the topology but on the metric itself. Thus, my problem is true if there exists a Borel compact metric on $X$ such that $f$ satisfies the given condition. – Raubeurght Hagh Apr 30 '16 at 14:20
  • In other words, Borel compact is not a well-defined notion of metrizable spaces, but only of metric spaces? – Raubeurght Hagh Apr 30 '16 at 14:24
  • Yes, as boundedness depends on the metric. – Henno Brandsma Apr 30 '16 at 14:26
  • Thank you very much. It was very helpful. I am grateful for your help. I wish you all the best. Sincerely, Raub. – Raubeurght Hagh Apr 30 '16 at 14:31
  • @RaubeurghtHagh You should note that the Heine-Borel property is also a property of metric spaces... – David C. Ullrich Apr 30 '16 at 14:35