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

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    $\begingroup$ Just take a Hausdorff space which is not regular, but is second countable. $\endgroup$
    – user87690
    Nov 19, 2014 at 13:00
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    $\begingroup$ For questions like this, the book 'Counterexamples in Topology' is very valuable! $\endgroup$
    – Hanno
    Nov 19, 2014 at 13:21
  • $\begingroup$ What have you tried? If you tell us this then we will be better able to help you. And it helps us feel that we are not just doing your homework for you. $\endgroup$
    – user1729
    Nov 19, 2014 at 13:26
  • $\begingroup$ @user1729 No its not homework problem. Actually I have studied till now countability axioms only and not seperation axioms till now. And examples of topological spaces I know doesn't fit with such thing.(R^2 dictionary order, co-finite. co-countable, lower limit, discrete, indiscrete) $\endgroup$
    – Sushil
    Nov 19, 2014 at 15:59
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    $\begingroup$ There is a search-engine (kind of) for topological space, where you can search for spaces with given properties, it is called $\pi$-base (topology.jdabbs.com). Your query would be the following: topology.jdabbs.com/search?q={%22and%22%3A[{%2227%22%3Atrue}%2C{%223%22%3Atrue}%2C{%2253%22%3Afalse}]} $\endgroup$
    – PhoemueX
    Nov 19, 2014 at 16:06

3 Answers 3

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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.)

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  • $\begingroup$ If a second countable space $X$ is regular then $X$ is metrizable. But how can we say that if $X$ is not regular then it is not metrizable ? $\endgroup$
    – Madhu
    Sep 11, 2018 at 7:13
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    $\begingroup$ @Madhu if it is metrizable, then it is regular, the proof is easy, use a small enough open ball $\endgroup$
    – Mirko
    Oct 28, 2018 at 18:54
  • $\begingroup$ @Mirko I do not understand why "$A$ is a closed subset that cannot be separated by disjoint neighborhoods from the origin", no open set around origin contains an element of $A$, in this topology. $\endgroup$
    – Tedebbur
    Jan 10, 2019 at 16:21
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    $\begingroup$ @Serpenche I mean separation as in the definition of regular topological spaces, and there is no such separation. That is, there are no disjoint open sets $P,Q$ such that the origin belongs to $P$ and such that $A$ is a subset of $Q$. $\endgroup$
    – Mirko
    Jan 23, 2019 at 15:50
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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.

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  • $\begingroup$ I cannot see, is it compact?, is there any second countable, non metrizable, compact hausdorff space? $\endgroup$ Jan 30, 2021 at 11:54
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    $\begingroup$ @Jale'dejaleuffnejale: It’s not compact: $L$ is an infinite, closed, discrete subset. Every second countable compact Hausdorff space is metrizable: this follows from the Urysohn metrization theorem. $\endgroup$ Jan 30, 2021 at 18:45
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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

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