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I have two questions related to metrizable spaces and countable network ;

  1. Can we find a continuous mapping from a separable metric space onto a non metrizable compact Hausdorff space.

  2. If a Hausdorff space $X$ has a countable network, then every compact subspace of X is metrizable.

The first one seem to be not true and the key is countable network, but not sure how to use it wisely.

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  • $\begingroup$ What do you mean by "network" in this context? $\endgroup$ – Robert Israel Dec 5 '14 at 18:28
  • $\begingroup$ @RobertIsrael Like a base, but its members need not be open. E.g. all singleton sets are always a network in any space. The network weigth $\operatorname{nw}(X)$ is the minimum of all sizes of networks for $X$. The image of a network is a network (which need not hold for bases). $\endgroup$ – Henno Brandsma Dec 5 '14 at 18:53
  • $\begingroup$ @RobertIsrael I just add link to countable network for reference. $\endgroup$ – henry Dec 5 '14 at 21:13
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Suppose $f$ is continuous onto from a separable metrisable $X$ to $Y$ which is compact Hausdorff. As $\operatorname{nw}(Y) = \operatorname{nw}(f[X]) \le \operatorname{nw}(X) = \aleph_0$, $Y$ has a countable network and so a countable base (as $Y$ is compact Hausdorff).

[Lemma: if $f: X \rightarrow Y$ is continuous and onto, and $\mathcal{N}$ is a network for $X$, then $\{f[N]: N \in \mathcal{N}\}$ is a network for $Y$. Proof: if $y \in Y$ and $y \in O \subseteq Y$, where $O$ is open, pick $x \in X$ with $f(x) = y$, and as $f^{-1}[O]$ is open and contains $x$, some $N \in \mathcal{N}$ exists with $x \in N \subseteq f^{-1}[O]$. But then $y \in f[N] \subseteq O$ and so indeed the images $f[N]$ form a network for $Y$. For bases this need not hold, as $f$ need not preserve open sets, but members of networks can be any type of set. This implies the $\operatorname{nw}(f[X]) \le \operatorname{nw}(X)$ fact.]

As to the second: if $X$ has a countable network and $K \subset X$ we have that $\operatorname{nw}(K) \le \operatorname{nw}(X) = \aleph_0$ and as $K$ is compact Hausdorff, it again has a countable base. (Intersecting all members of a network for $X$ with $K$ gives us a network for $K$, at most the size of the first network.)

We here use the standard fact (due to Arhangel'skij) that for compact Hausdorff spaces $X$: $\operatorname{nw}(X) = \operatorname{w}(X)$, plus the consequence of the Urysohn metrisation theorem that a second countable compact Hausdorff space is metrisable.

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