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I have learnt that if a regular, second countable space is normal(Theorem 32.1 of Munkres's Topology), and that a regular, second countable space is metrizable(Urysohn metrization theorem, Theorem 34.1). So, I want to analyize these four properties using a Venn diagram, and I would like to know if the diagram is correct, and examples for each part.

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Normality implies regularity, so there is no space that is normal but not regular. Also, metrizable space is always normal. I would like to know some examples each in 1, 2, 3, 4, 5 of the diagram. Here are some of my thoughts:

  1. $\mathbb{R}_l^2$(Sorgenfrey plane)
  2. $\mathbb{R}_l$(Sorgenfrey line)
  3. ??
  4. A good space, such as $\mathbb{R}$ with Euclidean metric topology
  5. ??
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2 Answers 2

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To expand on Sourav's excellent answer. You missed a crucial point in Uryshon Metrization theorem that the space must be Hausdorff.

1.An uncountable discrete space is metrizable given by the discrete metric. It is regular as given any point $x$ and a closed set $C$. we have $\{x\}$ and $C$ are disjoint open neighbourhoods of $\{x\}$ and $C$. Similarly it is normal as given two closed sets $F,C$ they themselves are open disjoint nbds of itself. Also these are $T_{1}$.

It is not second countable as if $\mathcal{B}$ be a basis for $(X,\tau_{\text{discrete}})$ then given an open set $\{x\}$ there exist a basic open set $B$ subset of $\{x\}$ and $x\in B\implies B=\{x\}$ . Thus $\{x\}\in\mathcal{B}\,,\forall\,x\in X$. Thus $\mathcal{B}$ is uncountable.

2 .The cofinite topology on $\Bbb{N}$ is not Hausdorff as given two points $x$ and $y$. Then any open set $U$ containing $x$ is of the form $U=\Bbb{N}\setminus F$ where $F$ is some finite subset of natural numbers. Then any open set containing $y$ say $V=\Bbb{N}\setminus C$ where $C$ is some finite set must have non-empty intersection with $U$. This is because $V\cap U= \Bbb{N}\setminus(F\cup C)$ which is infinite (because $F$ and $C$ were finite). In fact you could have worked with any infinite set.

Thus it is not metrizable as it is not Hausdorff .

This set is second countable as the set of all finite subsets of a countable set is countable. Hence you can just take $\{N\setminus F\}_{F\subset\Bbb{N}\,,\text{finite}}$ to be a countable basis.

This is a Normal space as each singleton is closed so it is $T_{1}$ . And given two disjoint closed sets $F,C$ . You have $F$,$C$ must be finite and hence $\Bbb{N}\setminus F$ and $\Bbb{N}\setminus C$ are disjoint open neighbourhoods. This also proves regularity as each singleton is clsoed.

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  • $\begingroup$ Now I see that I was missing a crucial point. Thanks for pointing it out. $\endgroup$
    – Joshua Woo
    May 20 at 10:43
  • $\begingroup$ @Mr.Gandalf Sauron Excellent explanation. $\endgroup$ May 21 at 13:56
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  1. $(X, \tau_{discrete}) $ where $X$ is uncountable.

Then $(X, \tau_{discrete}) $ is metrizable (generated by discere metric ), normal , regular but it is not second countable.

  1. $(\Bbb{N}, \tau_{cofinite}) $

Then it is second countable but not metrizable, regular, normal.

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  • $\begingroup$ Indiscrete spaces are regular and normal since there are not many closed sets around. $\endgroup$
    – Ruy
    May 20 at 4:43
  • $\begingroup$ @Ruy Hope it looks right now. $\endgroup$ May 20 at 5:07
  • $\begingroup$ Thank you for the answer. I have a question, why is the cofinite topology in $X$(uncountable) not metrizable? $\endgroup$
    – Joshua Woo
    May 20 at 6:50
  • $\begingroup$ @JoshuaWoo a metrizable space must satisfy $T_2$ axiom but cofinite topology on an infinite set is not Hausdorff. $\endgroup$ May 20 at 8:00
  • $\begingroup$ Ahh I've missed an important point. Thank you $\endgroup$
    – Joshua Woo
    May 20 at 10:43

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