We know that every metric space is normal. We know also that a normal, second countable space is metrizable.

What is an example of a normal space that is not metrizable?

Thanks for your help.

  • 1
    $\begingroup$ Think about a "large" compact Hausdorff space, for example $[0,1]^I$ for "large" $I$. $\endgroup$
    – martini
    Jun 13 '12 at 18:23
  • 1
    $\begingroup$ If you already know that normal+second countable = metrizable; try searching for normal which is not second-countable. $\endgroup$
    – Asaf Karagila
    Jun 13 '12 at 18:33
  • $\begingroup$ The Sorgenfrey Line is one example. $\endgroup$ Jun 13 '12 at 18:54
  • $\begingroup$ Second-countable normal spaces need not be metrizable: the trivial topology on a set of at least two elements is a counterexample. $\endgroup$ Jun 13 '12 at 19:00
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    $\begingroup$ @ChrisEagle: I guess it depends how you define normal. Some include Hausdorff in the definition. $\endgroup$
    – T. Eskin
    Jun 13 '12 at 19:29

Edit: In light of the comments, I thought it prudent to give the precise definition of normality used in the reference. For the list I've provided, a normal space $X$ is one satisfying:

  • T1 property: For all $x,y \in X$ there exist open sets $U_x$ and $U_y$ containing $x$ and $y$, respectively, such that $x \notin U_y$ and $y \notin U_x$.
  • If $A$ and $B$ are disjoint closed sets in $X$, there exist disjoint open sets $U_A$ and $U_B$ containing $A$ and $B$, respectively.

The following examples of normal, non-metrizable spaces come from $\pi$-Base, which is a searchable database inspired by Steen and Seebach's Counterexamples in Topology. You can learn more about each space by visiting the search result.

Alexandroff Square

Appert Space

Arens-Fort Space

Baire Product Metric on $\mathbb{R}^\omega$

Bing's Discrete Extension Space

Boolean Product Topology on $\mathbb{R}^\omega$

Closed Ordinal Space $[0,\Omega]$

Concentric Circles

Countable Excluded Point Topology

Deleted Integer Topology

Divisor Topology

Either-Or Topology

Finite Excluded Point Topology

Fortissimo Space

Helly Space

Hjalmar Ekdal Topology


Lexicographic Ordering on the Unit Square

Michael's Closed Subspace

Nested Interval Topology

Odd-Even Topology

One Point Compactification Topology

One-point Lindelofication of $\omega_1$

Open Ordinal Space $[0,\Omega)$

Radial Interval Topology

Right Half-Open Interval Topology

Right Order Topology on $\mathbb{R}$

Rudin's Dowker Space

Sierpinski Space

Single Ultrafilter Topology

Stone-Cech Compactification of the Integers

The Extended Long Line

The Integer Broom

The Long Line

Tychonoff Plank

Uncountable Excluded Point Topology

Uncountable Fort Space


Rather than list specific spaces, I thought that I’d mention a few classes of spaces.

A compact metric space has cardinality at most $2^\omega$, so every compact Hausdorff space of cardinality greater than $2^\omega$ is normal and non-metrizable. In particular, this includes every product of compact Hausdorff spaces with at least $2^\omega$ non-trivial factors.

Let $I$ be an index set, for each $i\in I$ let $X_i$ be a space with at least two points, and for each $i\in I$ let $p_i\in X_i$. Let $$X=\left\{x\in\prod_{i\in I}X_i:|\{i\in I:x_i\ne p_i\}|\le\omega\right\}$$ as a subspace of the Tikhonov product of the $X_i$; such spaces are called $\Sigma$-products. If $I$ is countable, the $\Sigma$-product is just the ordinary Tikhonov product, but if $I$ is uncountable it’s something new.

Proposition: If $I$ is uncountable and each $X_i$ is $T_1$, $X$ is not paracompact (and therefore not metrizable).

Proof: Let $I_0=\{i_\xi:\xi<\omega_1\}$ be a subset of $I$ of cardinality $\omega_1$, and for each $\xi<\omega_1$ fix $q_{i_\xi}\in X_{i_\xi}\setminus\{p_{i_\xi}\}$. For $\eta<\omega_1$ define $x^\eta\in X$ by $$x^\eta_i=\begin{cases}q_{i_\xi},&\text{if }i=i_\xi\text{ and }\xi<\eta\\p_i,&\text{otherwise}\;.\end{cases}$$ It’s not hard to check that $\{x^\eta:\eta<\omega_1\}$ is a closed subspace of $X$ homeomorphic to $\omega_1$ (with the order topology), which is not paracompact. $\dashv$

However, it’s a theorem of Mary Ellen Rudin and, independently, S.P. Gul’ko that $\Sigma$-products of metric spaces are always normal. (The proof is highly non-trivial and can be found in Teodor C. Przymusiński, Products of Normal Spaces, in the Handbook of Set-Theoretic Topology, K. Kunen & J.E. Vaughan, eds., where the result is Theorem 7.4.) Thus, every uncountable $\Sigma$-product of non-trivial metric spaces is an example of a normal, non-metrizable space.

It’s well known that every linearly ordered space [LOTS] is hereditarily normal. This means that every generalized ordered [GO] space is hereditarily normal, since the GO-spaces are precisely the subspaces of linearly ordered spaces. (The actual definition of a GO-space is that it’s a space $X$ equipped with a linear order $\le$ whose topology has a base consisting of $\le$-intervals, not necessarily open, but this is equivalent to being a subspace of a LOTS. An example is the Sorgenfrey line.)

Thus, any non-metrizable GO-space is an example. The most straightforward way for a GO-space to fail to be metrizable is to have a point with uncountable character, i.e., a point that has no countable local base; this automatically includes all ordinal spaces $\alpha$ for $\alpha>\omega_1$ and many of their subspaces. This is far from the only way, of course. The Sorgenfrey line, for example, is first countable but fails to be metrizable for a variety of reasons: it’s separable and Lindelöf but not second countable, and its square is neither normal nor Lindelöf. It’s rather easy to come up with all sorts of variations on this theme.


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