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I want to know some examples of topological spaces which are not metrizable. Of course one can construct a lot of such spaces but what I am looking for really is spaces which are important in other areas of mathematics like analysis or algebra. I know most spaces arising naturally in other areas of mathematics are metrizable because of the Urysohn metrization theorem. But still there must be some examples of non-metrizable spaces.So far I know the following examples:

  1. Zariski topology
  2. Weak* topology on $X^{*}$ if X is an infinite dimensional Banach space
  3. The topological vector space of all functions $f:\mathbb{R}\rightarrow\mathbb{R}\ \ $ under pointwise convergence.

Your help is appreciated.

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Have you had a look at Counterexamples in Topology? – Nate Eldredge Jan 14 '11 at 5:31
@Anthony: No, that's not true. It is true that the unit ball in $X^{\ast}$ is weak$^{\ast}$-metrizable if $X$ is separable, but this does by no means imply that the weak$^{\ast}$-topology on all of $X^{\ast}$ is metrizable. In fact, the weak$^{\ast}$-topology is not even first countable if $X$ is infinite-dimensional. – Theo Buehler Jan 14 '11 at 7:05
Ricky, can I encourage you to delete your comment? That's unnecessarily aggressive language, I would say. Apart from anything else, Sudip is new to MathOverflow. – Tom Leinster Jan 14 '11 at 12:42
I must say I agree with Tom Leinster. Furthermore, I have real questions about using votes to bargain or in a coercive manner. Perhaps someone should take this to meta? – drbobmeister Jan 14 '11 at 18:13
Ricky, you can remove your upvote if you choose to. I am here to learn the answer to my question, not to get votes. – Sudip Paul Jan 14 '11 at 22:07

13 Answers 13

up vote 9 down vote accepted

I'm not an expert, but I believe the space of Distributions (in any number of variables), (a.k.a. generalised functions, including the Dirac delta, its derivatives, etc.) as used in PDE theory, is a topological vector space, but non-metrisable; even though sequences are sufficient to do everything.

However, the subspace of tempered (Schwartz) distributions, as used in Fourier Analysis, is metrisable; it is a Fréchet space.

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The pre-dual, that is the space $\mathcal{D}(\Omega)$ of $C^\infty$ compactly supported functions on $\Omega$, is already non-metrizable (it is a countable union of closed linear subspaces, namely, the subspaces of functions with supports in given compact sets. It is sequentially complete, so it can't be metrizable by Baire theorem). Yet it shares several properties of metric spaces, since it is an inductive limit of Fréchet spaces; which is the reason why sequences suffice. – Pietro Majer Jan 14 '11 at 7:24
Oddly, this false statement seems to have stood the test of time. The space of tempered distributions is NOT metric although, being a Silva space, i.e. an inductive limit of a sequence of Banach spaces with compact intertwining maps it shares many of their properties (see, e.g., Köthe, "Topological linear spaces". Functional analysis abounds in important non-metrisable spaces, in distrubtion theory as mentioned above, but also in measure theory. – barcelos Mar 27 '14 at 6:24

From my perspective as an analyst, non-metrizable spaces usually arise for one of the following reasons:

  1. Separation axiom failure: the space is not, e.g., normal. This mostly happens when the space is not even Hausdorff (spaces that are Hausdorff but not normal are usually too exotic to arise much). Often this is for simple reasons; for example, the topology is defined by a pseudometric such as a seminorm. In this case we usually mod out by zero-distance pairs and try again.

  2. The space is too big: examples like the uncountable ordinal and the long line fall under this heading. But they are still locally metrizable, so we usually allow them if we're trying to prove local theorems, but forbid them for global statements.

  3. The topology is too weak, and usually not even first countable, so sequences are not enough to define the topology. Most of these examples are based on a product or pointwise-convergence topology: the weak-* topology on the dual $X^\*$ of a Banach space $X$, the spectrum of a $C^\*$-algebra, Stone-Čech compactifications, etc. (And actually, in the first case, it is often enough just to look at the unit ball of $X^*$, which is metrizable if $X$ is separable, which in applications it usually is.) As compensation, some corollary of Tychonoff's theorem gives us some compactness, which is probably the only reason we've agreed to put up with such an annoyingly weak topology in the first place.

  4. Someone is abusing the language of topology, e.g. Fürstenberg's "topological" proof of the infinitude of the primes.

  5. I've wandered into an algebraic geometry seminar by mistake. Grouchier analysts may consider this to fall under the previous heading. ;-)

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re 1, or just: the space is not perfectly normal Hausdorff. – Ricky Demer Jan 14 '11 at 6:14
Could you elaborate on point 4? I don't think I get the point because the proof involves a perfectly honest topology on the integers. Do you mean to say that you don't need the language of topology in order to grasp the essence of the proof? – Theo Buehler Jan 14 '11 at 7:24
Theo: Even so, #4 makes no sense. The profinite topology on the integers is not metrizable whether or not you need it in a proof. I would suggest replacing #4 with 'someone is making instructive counterexamples' or something similar, cf. Counterexamples in Topology. – Ketil Tveiten Jan 14 '11 at 9:52
I believe to remember, that the arithmetic sequences are chosen as a basis for closed sets (in 4.). The reason, why this space is not metrizable is not among the reasons mentioned before (btw. what is it??). So 4 is perfectly fine. – HenrikRüping Jan 14 '11 at 13:12

Every finite topological space which is not discrete is not Hausdorff and hence not metrizable. Yet, there are lots of finite topological spaces: they are in one-to-one correspondence with finite preordered sets.

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This generalizes to infinite preorders and Alexandroff spaces: – Michał Kukieła Jan 14 '11 at 14:37

Because nobody has mentioned it, and in my mind, this is one of the biggest surprises set-theoretic topology has ever given mathematics, here it is:

The following is independent of ZFC: Every normal Moore space is metrizable.


Technically speaking, a Moore space is a developable regular Hausdorff space. To see that Moore spaces can occur in nature, and kind of get a feel for what they are consider my answer to this question Topological Rings .

Some Results

  • Assuming $CH$, there is a normal Moore space which fails to be metrizable.

Reference: William G. Fleissner's "Normal nonmetrizable Moore space from continuum hypothesis or nonexistence of inner models with measurable cardinals"

  • If every normal Moore space is metrizable, then there exists an inner model with a measurable cardinal

Reference: William G. Fleissner's "If All Normal Moore Spaces Are Metrizable, Then There Is An Inner Model With A Measurable Cardinal"

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The strong Whitney topology on the space of smooth maps $C^{\infty}(M;N)$ between two manifolds is not metrisable unless $M$ is compact (this is closely related to Thomas Roths answer). Inside this space, there is the subspace of functions transverse to a given submanifold $W \subset N$. This subspace is dense, but not sequentially dense, which led me to confusion when I learnt differential topology.

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Any locally infinite simplicial complex.

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Like the curve complex! (Go on, admit that's what you were thinking of.) – HJRW Jan 14 '11 at 20:59
Actually, I was thinking of the complex of curves. ;) – Richard Kent Jan 14 '11 at 21:09

Functional analysis, and in particular the theory of operator algebras, involves many such topologies. For example, neither the strong operator nor the weak operator topology on $B(H)$, the algebra of bounded operators on a separable Hilbert space $H$ is metrizable (but give metrizable topologies when restricted to bounded sets). See the Wikipedia article for more such examples.

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Another non-metrizable space which is common in algebraic topology is $\mathbb{R}^{\infty}$ with the colimit topology. A subset $U$ is open iff the intersection with any finite $\mathbb{R}^n$ is open in the usual topology. This is a locally convex topological vector space, but its properties seem rather ugly to me.

Claim: $\mathbb{R}^{\infty}$ is not metrizable.

Proof: Suppose that there is a metric $d$ on $\mathbb{R}^{\infty}$ that defines the topology. Let $e_k$ be the $k$-th basis vector. There exists a sequence $t_k \in (0, \infty)$ such that $d(0,t_k e_k)$ converges to zero. Such a sequence can be constructed as follows. If $d(0,e_k) \leq 1/k$, take $t_k=1$. Otherwise, the function $d_k:[0,1] \to \mathbb{R}$, $t \mapsto d(0,t e_k)$ is continuous, $d_k(0)=0$, whence there exists $t_k$ with $d(t_k)=1/k$. The sequence $v_k:=t_k e_k$ converges to $0$ in $\mathbb{R}^{\infty}$. Let $f:\mathbb{R}^{\infty} \to \mathbb{R}$ be the linear function given by $f(e_k)=1/t_k$. The restriction of $f$ to any $\mathbb{R}^n \subset \mathbb{R}^{\infty}$ is clearly continuous and so by the definition of the colimit topology, $f$ is continuous (in fact any linear map from $\mathbb{R}^{\infty}$ to a topological vector space is continuous). In particular, if $v_k \to 0$ in $\mathbb{R}^{\infty}$, then $f(v_k)\to 0$. But the sequence $v_k$ constructed above satisfies $f(v_k)=1$, contradiction.

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Function spaces are sometimes not metrizable. Let $X$ and $Y$ be topological spaces, and $C(X,Y)$ be the space of continuous maps from $X$ to $Y$, topologized in the compact open topology. Then $C(X,Y)$ need not be metrizable (it is if $X$ is a compact, and $Y$ is a metric space, it is though).

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I like the example of the Double Arrow space. Take the set $[0,1] \times \{0,1\}$ with the lexicographic order. Then with the order topology this space is compact, separable but not metrizable.

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Sorgenfrey line is not metrizable, since it is separable but does not satisfy the second axiom of countability.

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Would you kindly remind us the definition? – Claudio Gorodski Mar 27 '14 at 2:35
The Sorgenfrey line or lower limit topology is $\mathbb{R}$ with a basis of $\lbrace [a,b)\rbrace$. I don't know of a particular use other than as a counterexample or test case. IIRC, describing the compact sets is a nice exercise. – Douglas Zare Mar 27 '14 at 9:20
A good reference is Topology, a first course by Munkres. See page 192 (1st edition) where it is shown that the Sorgenfrey line is separable (has a countable dense subset) but is not second countable. On the other hand, we know that a separable metric space is second countable:… – Todd Trimble Mar 27 '14 at 10:53

A particularly nice example that occurred naturally in my research is as follows:

Let $A$ algebra of $L^\infty$-equivalence classes of piecewise continuous functions on $[0,1]$ (or equivalently the algebra of PC functions with $f(x)$ equal to the right hand limit of $f$ at $x$ for each $x$) equipped with the uniform norm. Take our topological space $X$ to be the maximal ideal space/character space of $A$ (put the weak star topology on the set of multiplicative linear functionals).

We get the following nice properties:

$X$ is compact and Hausdorff (since it's the maximal ideal space of a commutative C*-algebra).

It's not metrizable because $A$ isn't separable; that metrizabilty of $X$ is equivalent to separability of $C(X)$ is a fun little exercise.

It's not too huge. The multiplicative linear functionals are just left limits, $x_l$, and right limits, $x_r$, at points $x$ in the interval so it has the same cardinality as $\mathbb R$.

$X$ is separable; let if $(x_n)$ are rationals tending to $x$ from the left then then ${x_n}_l$ tends to $x_l$ and likewise on the right.

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One can also ask for a weaker form of metrizability, where distances are measured in any ordered field (not neccesarily the reals). A example for such a bigger ordered field is the hyperreals. Lets call them $\mathbb{R}^*$.

For example the hyperreals are not $\mathbb{R}$-metrizable, but they are $\mathbb{R}^*$-metrizable just by setting $d(a,b):=|a-b|$. So we have already a example for a nonmetrizable space.

Given any ordered field $F$ and a $F$-metrizable space $S$. Then one can find a local basis at every point $s\in S$ via $\{B_\varepsilon(s)| \varepsilon \in F,\varepsilon >0\}$. Note that this basis is totally ordered (under the inclusion).

Now consider the space $X:=\prod_{i\in I} \{0;1\}$, where $I$ is a uncountable set and let $x\in X$ be any point. One can show, that there is no local basis at $x$, that is totally ordered. So this space is not metrizable for any ordered field $F$.

I think this is quite surprising. Heuristically speaking, this space is too big to be $\mathbb{R}$-metrizable. But it is even not $F$ - metrizable for bigger $F$'s.

It shows, that Urysohns metrization theorem cannot be generalized to $\mathbb{R}^*$-metrics, i.e. a statement like

"A regular Hausdorff space with a basis for the topology of cardinality $\le$ ? is $ \mathbb{R}^*$-metrizable. "

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