Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Think about a "large" compact Hausdorff space, for example $[0,1]^I$ for "large" $I$. –  martini Jun 13 '12 at 18:23
    
If you already know that normal+second countable = metrizable; try searching for normal which is not second-countable. –  Asaf Karagila Jun 13 '12 at 18:33
    
The Sorgenfrey Line is one example. –  David Mitra Jun 13 '12 at 18:54
    
Second-countable normal spaces need not be metrizable: the trivial topology on a set of at least two elements is a counterexample. –  Chris Eagle Jun 13 '12 at 19:00
2  
@ChrisEagle: I guess it depends how you define normal. Some include Hausdorff in the definition. –  Thomas E. Jun 13 '12 at 19:29
add comment

2 Answers

up vote 10 down vote accepted

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 Spacebook, which is a searchable version of the data from Counterexamples in Topology. You can find more about most of these spaces on the web (and I will be happy to elaborate on any you cannot find).

Alexandroff Square
Appert Space
Arens-Fort Space
Bing’s Discrete Extension Space
Closed Uncountable Ordinal Space
Concentric Circles
Discrete Irrational Extension of the Reals
Fortissimo Space
Helly Space
Lexicographic Ordering on the Unit Square
Michael’s Closed Subspace
Open Uncountable Ordinal Space
Radial Interval Topology
Right Half-Open Interval Topology
Single Ultrafilter Topology
Stone-Cech Compactification of the Integers
The Extended Long Line
The Long Line
Tychonoff Plank
Uncountable Cartesian Product of Unit Interval
Uncountable Fort Space
Weak Parallel Line Topology

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.