Examples of topological spaces satisfying certain properties

Question

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.

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. ??

Answer 1

3. $(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.

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

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

Answer 2

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.