# Example of Hausdorff and Second Countable Space that is Not Metrizable

Does there exist topological space that is Hausdorff and second countable but not metrizable?

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$\pi$-Base is a database of topological spaces inspired by Steen and Seebach’s Counterexamples in Topology. It lists the following fourteen second countable, Hausdorff spaces that are not metrizable. You can learn more about an of them from the search result.

Arens Square

Double Origin Topology

Indiscrete Irrational Extension of $\mathbb{R}$

Indiscrete Rational Extension of $\mathbb{R}$

Irrational Slope Topology

Irregular Lattice Topology

Minimal Hausdorff Topology

Pointed Irrational Extension of $\mathbb{R}$

Prime Integer Topology

Relatively Prime Integer Topology

Roy’s Lattice Space

Roy’s Lattice Subspace

Simplified Arens Square

Smirnov’s Deleted Sequence Topology

Here is one of the simpler examples. Let

$$A=\left\{\left\langle\frac1m,\frac1n\right\rangle:m,n\in\Bbb Z^+\right\}$$

and

$$L=\left\{\left\langle\frac1m,0\right\rangle:m\in\Bbb Z^+\right\}\;.$$

Let $X=\{\langle 0,0\rangle\}\cup L\cup A$. These ordered pairs are a bit clumsy, so let me introduce some abbreviations: for $m,n\in\Bbb Z^+$ let $a_{m,n}=\left\langle\frac1m,\frac1n\right\rangle$, for $m\in\Bbb Z^+$ let $x_m=\left\langle\frac1m,0\right\rangle$, and let $p=\langle 0,0\rangle$.

We topologize $X$ as follows.

• Points of $A$ are isolated; i.e., $\{a_{m,n}\}$ is an open set for each $m,n\in\Bbb Z^+$.
• For $m,n\in\Bbb Z^+$ let $B_n(x_m)=\{x_m\}\cup\{a_{m,k}:k\ge n\}$; $\{B_n(x_m):n\in\Bbb Z^+\}$ is a local base of open nbhds of $x_m$.

So far this simply gives $A\cup L$ the topology that it inherits from the usual topology on $\Bbb R^2$. It’s only when we define the topology at $p$ that we do something different. First, though, for $m\in\Bbb Z^+$ let $A_m=\{a_{m,n}:n\in\Bbb Z^+\}$; $Y_m$ is the set of points of $A$ lying on the line $y=\frac1m$ in the plane.

• For $n\in\Bbb Z^+$ let $B_n(p)=\{p\}\cup\bigcup_{k\ge n}A_k$; $\{B_n(p):n\in\Bbb Z^+\}$ is a local base of open nbhds of $p$.

Clearly $X$ is first countable, and it’s not hard to check that it’s Hausdorff as well. For example, $B_{n+1}(p)$ and $B_1(x_n)$ are disjoint open nbhds of $p$ and $x_n$, respectively. But $X$ isn’t regular, so it can’t be metrizable. Specifically, the set $L$ is closed, and $p\notin L$, but $p$ and $L$ do not have disjoint open nbhds.

Let $X$ be the real line with the topology in which the usual open sets are open, and in addition $U\setminus A$ is open for any $U$ that is open in the usual topology, where $A=\{\frac1n : n=1,2,…\}$. In other words, every point except the origin has its usual neighborhoods, and the basic neighborhoods of the origin are of the form $(-\varepsilon,\varepsilon)\setminus A$. Then $A$ is a closed subspace that cannot be separated by disjoint neighborhoods from the origin, so $X$ is not regular, and not metrizable. $X$ is Hausdorff since its topology is stronger than the usual topology which is Hausdorff. If we take a countable basis for the usual topology together with a countable local basis at the origin for the new topology, then these two together form a countable basis for the new topology. (This is a standard example, e.g. Ex.1.5.6 in General Topology by R. Engelking.)