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I know compact Hausdorff space is normal, and metric space is normal.

What about compact normal space? Is it a metric space?

My naive guess is that normality is nothing to do with metric and compactness(finiteness) is also nothing to do with metric so it seems it is not a metric space. But how one can prove or disprove this more than my idea?

Arctic Char
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phy_math
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1 Answers1

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A beautiful example to the contrary is the Double Arrow Space, one of the classics in topology.

It's compact, perfectly normal ($T_6$), hereditarily Lindelöf, hereditarily separable, first countable, but not metrisable as it does not have a countable base.

Henno Brandsma
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  • Another elementary example is the lexicographic-order topology on $X=[0,1]^2$. All linear spaces are normal, and $X$ has the $lub$ property (every subset has a $lub$) so $X$ is compact. A compact metrisable space is separable, but ${{r}\times (1/3,2/3):r\in [0,1]}$ is an uncountable family of pairwise-disjoint open subsets of $X,$ so $X$ is not separable. – DanielWainfleet Jul 26 '21 at 14:56
  • @DanielWainfleet double arrow is a subspace of that square. But it’s nicer ( hereditarily separable/ Lindelöf). – Henno Brandsma Jul 26 '21 at 14:58
  • Yes. The lex. square is not $T_6$, among other things. The Double-Arrow is a great example of how far you might go and still not be metrisable. – DanielWainfleet Jul 26 '21 at 15:11