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I have a little doubt about compactness in metric spaces. I have this homework where I have to prove that $[0,1]^\omega$ with the uniform topology is not countably compact. As a consequence of Thychonoff's theorem $[0,1]^\omega$ is compact and it is also metrizable for being the topology induced by the uniform metric. My problem is that, as far as I know, compactness and countably compactness are equivalent in metric spaces. The latter means that $[0,1]^\omega$ should be countably compact (but it is not). I know this might be a silly question but I really don't understand. Can someone tell me were am I wrong? Thanks in advance

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    $\begingroup$ Tychonoff's Theorem says $[0,1]^\omega$ is compact in the product topology. (Which, apparently, is different from the topology induced by the uniform metric.) $\endgroup$ – David Mitra Nov 1 '18 at 3:32
  • $\begingroup$ The product topology in the Tychonoff Theorem is also called the Tychonofg product topology to distinguish it from other topologies such as the box product topology $\endgroup$ – DanielWainfleet Nov 1 '18 at 19:21
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Tychonoff's theorem says that $[0,1]^\omega$ is compact in the product topology, which is in this case also induced by a metric (e.g. $d(x,y) = \sum_{n=1}^\infty \frac{1}{2^n} |x_n -y_n|$ will do; this topology is thus also countably compact, as this is equivalent to compactness in metrisable spaces) but this topology is different from the topology induced by the uniform metric (what Munkres calls (confusingly, IMHO) the uniform topology), so Tychonoff's theorem has no bearing on this.

The fact that the uniform topology is strictly finer than the (compact Hausdorff) product topology already implies it's not compact (and thus not countably compact), but a directer argument is available too:the set $A = \{x\in [0,1]^\omega: \forall n: x_n \in \{0,1\}\}$ is (uncountable and) closed and discrete in the uniform metric. This implies that $[0,1]^\omega$ in that topology also is not (countably) compact. It is true that countable compactness and compactness are equivalent in metrisable spaces (so in the uniform topology too) but both are easy to disprove in this case, so it doesn't help nor hinder.

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  • $\begingroup$ To the proposer, re the 2nd paragraph, 1st sentence: It is fairly easy to prove that if $X, Y$ are compact Hausdorff spaces and $f:X\to Y$ is a continuous bijection then $f$ is a homeomorphism. So if $T, T'$ are compact Hausdorff topologies on $X$ with $T\subset T'$ then $id_X$ is a continuous bijection from $(X, T')$ to $(X,T)$ so $ id_X$ is a homeomorphism, So the image $id_X(t')\in T$ for all $t'\in T'$ so (since $T\subset T'$) we have $T=T'. $ $\endgroup$ – DanielWainfleet Nov 1 '18 at 19:31

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