In introductory real analysis, I dealt only with $\mathbb{R}^n$. Then I saw that limits can be defined in more abstract spaces than $\mathbb{R}^n$, namely the metric spaces. This abstraction seemed "natural"to me. Then, I knew the topological spaces. However, this time the abstraction did not seem natural/useful to me. Then one runs into problems of classifying spaces into normal/ first countable ... In my opinion, these resulted from the high level of abstraction adopted by studying topological spaces. When one uses a more general definition for a space, it is possible that the number of uninteresting objects increase. I guess this is what happened here, we use a very general definition for topological spaces, we get a lot of uninteresting spaces, then we go back and make classifications such as normal, Hausdorff,..

I was trying to justify to myself why are topological spaces are good to study. The best and only reason I can propose is that the category $Top$ is bicomplete.

Question 1: (Alternatives to $Top$)

If this is the only reason, can't there exist a "smaller" category such that it contains all metric spaces and is bicomplete ?

Question 2: (History of topological spaces)

I mentioned that the abstraction from metric spaces to topological spaces does not seem very natural to me. I suspect that historically, metric spaces were studied before topological spaces. If this is the case, I'd like to know what was the motivation/justification for this abstraction.

Question 3: (Applications of non-metric topology outside topology)

I mentioned earlier that "we get a lot of uninteresting spaces". Perhaps I am wrong (I hope I am wrong). I would value non-metric topological spaces more, if I see examples of theorems such that:

1) The theorems are in a branch of mathematics outside Topology

2) The theorems are proven with the aid of topology

3) The topological part about the proof of the theorem is about a non-metric space

Edit: ${}$ non-artificial instances of non-metric spaces appearing in other branches of math are valuable as well.

Thank you

  • 7
    $\begingroup$ Have you studied things like algebraic topology and cohomology yet? There are cases where you want to study things that are only dependent on how the space is connected. Even though a metric may still exist, dragging it along when it isn't "respected" by anything you're doing would only make things more difficult. Topology cleanly isolates the important features in such problems. In particular, I take some issue with point 3: the metric isn't automatically useful or relevant just because it's there. $\endgroup$ Commented Jul 31, 2013 at 11:54
  • 14
    $\begingroup$ "(Un)Interesting" is somewhat subjective. Let's say that there are important topologies which are not metrizable and not even Hausdorff. The first example that comes to mind is the Zariski topology. en.wikipedia.org/wiki/Zariski_topology $\endgroup$ Commented Jul 31, 2013 at 11:55
  • 11
    $\begingroup$ Im just going to quote Munkres in his book, Topology: "The definition of a topological space that is now standard was a long time in being formulated. Various mathematicians - Frechét, Hausdorff, and others - proposed different definitions over a period of years during the first decades of the twentieth century, but it took quite a while before mathematicians settled on the one that seemed most suitable. They wanted, of course, a definition that was as broad as possible, so that it would include as special cases all the various examples that were useful in mathematics - euclidean space,... $\endgroup$
    – Integral
    Commented Jul 31, 2013 at 12:14
  • 11
    $\begingroup$ ..., infinite-dimensional euclidean space, and function spaces among them - but they also wanted the definition to be narrow enough that the standard theorems about these familiar spaces would hold for topological spaces in general. This is always the problem when one is trying to formulate a new mathematical concept, to decide how general its definition should be. The definition finally settled on may seem a bit abstract, but as you work through the various ways of constructing topological spaces, you will get a better feeling for what the concept means.” $\endgroup$
    – Integral
    Commented Jul 31, 2013 at 12:15
  • 4
    $\begingroup$ You might be interested to look into the so-called compactness theorem, which is about model theory, but which, as its name suggests, arises from a topological consideration of the space of possible theories, a Stone space, which is a product of compact spaces and therefore compact as a consequence of Tychonoff's theorem. The compactness theorem says that a system of axioms is consistent if and only if every finite subset is consistent. $\endgroup$
    – MJD
    Commented Jul 31, 2013 at 14:22

7 Answers 7


The space $\mathbb{R}^\mathbb{R}$, which is the space of all functions from $\mathbb{R}$ to itself with the topology of pointwise convergence, is not a metric space (it is not even first countable). This kind of function space arises in many areas of math. The issue is that only countable products of metric spaces need to be metric, but function spaces like this are uncountable products.


This addresses only the third of your questions.

There has been a lot of interesting work lately by Martín Escardó, Paul Taylor, and others that interprets computability in terms of topology. It turns out, for example, that a predicate function $p:X\to\def\Bool{\mathbf{Bool}}\Bool$ is effectively computable if and only if it is continuous. But to make it work you cannot give $\Bool = \{\mathbf{True}, \mathbf{False}\}$ a metric topology. Rather, the correct topology is the Sierpiński topology in which $\{\mathbf{True}\}$ is open and $\{\mathbf{False}\}$ is not.

Other topological notions turn out to be important. For example, consider the function $\def\fa{\mathtt{forall}}\fa$, which takes a computable predicate $p: X\to\Bool$, and returns the truth of $$\forall x\in X. p(x).$$ ($\fa$ is a mapping from $\Bool^X\to\Bool$.) It transpires that $\fa$ is computable if and only if $X$ is topologically compact. This has some weird-seeming implications: $\fa$ is not guaranteed to terminate on the natural numbers $\Bbb N$, but it is computable and guaranteed to terminate on the Alexandroff compactification $\Bbb N\cup \{+\infty\}$. And similarly since the Cantor set of all sequences $\Bbb N\to \Bool$ is compact, $\fa$ can be effectively implemented to give a correct result for any predicate $p$ defined on sequences, even though the space of sequences is uncountable.

  • 2
    $\begingroup$ I don't know a lot of logic, but this seems to be an interesting application +1 $\endgroup$
    – Amr
    Commented Jul 31, 2013 at 21:59
  • 2
    $\begingroup$ This doesn't sound quite right. $Bool$ ought to be the discrete space with two points. In cs.bham.ac.uk/~mhe/papers/barbados.pdf , Escardo defines the sierpinski space as its own type (distinct from $Bool$). $\endgroup$
    – Tac-Tics
    Commented Aug 6, 2013 at 19:59
  • $\begingroup$ Yes, thanks for pointing this out. I was conflating (only partly on purpose) the Sierpiński space, which has two elements, with $\mathbf{Bool}$, which has three. If I can think of a good way to rewrite this to make the distinction clear, without excessively complicating the discussion, I will come back and fix it. $\endgroup$
    – MJD
    Commented Aug 7, 2013 at 13:28
  • $\begingroup$ I think I may have overstated the relationship between the computability of forall and topological compactness; I think the implication is only one way. But I'm still working through the papers and I'm not completely sure. $\endgroup$
    – MJD
    Commented Dec 31, 2013 at 4:32

QUOTE:non-artificial instances of non-metric spaces appearing in other branches of math are valuable as well.

Let $U$ be a non-empty open subset of $\mathbb{R}^n$, $n \ge 1$. A test function on $U$ is a smooth function $U \rightarrow \mathbb{R}$ with compact support. Let $D(U)$ be the set of test functions on $U$. This is a vector space over $\mathbb{R}$. We can define a certain topology on $D(U)$ which makes $D(U)$ locally convex and complete. This topology is not metrizable. The dual space of $D(U)$ is called the space of distributions on $U$.



Most answers focus on the third question, so I'll try to say something about the first and the second. I think metric spaces (with non-expanding maps) are not suitable for topology. The reason is metric spaces do not glue well enough.

Let $A$ and $B$ be metric spaces, and $C$ be a subspace of both (that is, there are isometries from $C$ to subspaces of $A$ and $B$). Is there a metric space which glues $A$ and $B$ along $C$? That's quite natural thing to do. Topologists love to glue spaces together. What is a sphere? It is two hemispheres (that is, disks) glued along the equator. Manifolds are spaces that could be glued from pieces of $\mathbb{R}^n$ along open subsets. And so on.

Well, let's try to build such a metric space. Take a union of $A$ and $B$ as sets and identify points "from $C$" in both. Then you need to specify a metric. How to measure distance between a point $x$ from $A$ and a point $y$ from $B$? There was no way to estimate it before glueing. Probably the only reasonable thing to try is to set $d(x, y)$ to be infimum of $d(x, z) + d(z', y)$ where $z$ is a point of $A$ and $z'$ is a point of $B$ which glues to $z$.

That would be nice, but here is a problem: if $x$ and $x'$ are two points of $A$, it may become quicker to go from $x$ to some point of $B$, travel in $B$, and then return to $A$ to $x'$ than to go from $x$ to $x'$ directly. The language of travelling is not mathematically meaningful here, it just means that there could exist a point $y$ in $B$ such that $d(x, y) + d(y, x')$ is less than $d(x, x')$. That happens in almost all cases.

It could be fixed. Well, you need to acknowledge the possibility of "going to $B$ and returning to $A$" several times, so in the end your formula for distance between two points would look like this: for all sequences of points $s_1,\ldots, s_k$ evaluate $d(x, s_1) + d(s_1, s_2) + \ldots + d(s_k, y)$ and take the infimum (for all $k$ and for all sequences).

It's also possible the infimum will be zero. You have no choice but to quotient such pairs of points together. I believe after these manipulations you get a correct "glueing" (that is, pushout) in the category of metric spaces with non-expanding maps, so the category isn't absolutely unreasonable. Let's put the issue of zero infimum aside for now.

Now what's wrong with this construction? It contradicts my intuition about what space means. I expect that if I attach something to a space and then ignore the added part, I get the original space back. It doesn't happen here: there is no way to recover metric on $A$ from that infimum-of-sums formula. $A$ maps into "glued" space, and it's even a monomorphism, but $A$ is not a part of glued space.

If you want $A$ to be a part of glued space, you need to identify $A$ with its image. So the objects in the category would be "sets with many equivalent metrics on them". Then the morphisms also have to change: morphism from $X$ to $Y$ is a mapping of underlying sets which becomes non-expanding map for some choices of metrics on the $X$ and $Y$ equivalent to given ones. Translating this condition to the language of one chosen metric leads to the usual epsilon-delta definition of continuous maps.

As soon as you get continuous maps, topological spaces are not far ahead. Thinking in epsilon-delta terms is almost the same as thinking about open sets. The definition of a topological space is more or less the most general thing for which there is a notion of continuity. The modern definition is surely not the first one, but all previous were quite close.

It was not produced while studying metric spaces, but rather when trying to formalize glueing spaces together and quotienting by subspaces. At least that's my interpretation: historically there were a lot of theorems about spaces glued from polyhedrons, and later for CW-complexes, so the convenience of glueing is absolutely necessary for the definition of a space.

  • $\begingroup$ Hi. This is why I said the best reason I can propose is that the category $Top$ is bicomplete. Now if this really the best reason, what would be the answer to question 1 (the one mentioned in my post) ? Thank you $\endgroup$
    – Amr
    Commented Jan 5, 2014 at 0:13
  • $\begingroup$ How is it "almost always the case" that $d(x,y) + d(y,x') < d(x,x')$ for some $y$? If we have isometric copies of $C$ in both spaces, we have $d(x,y)+d(y,x) = d(x,z)+d(z,y)+d(y,z')+d(z',x')$ for some $z,z'$ in $C$ (to ignore the fact that it's an infimum, take some $z$ and $z'$ making this almost an equality). Then this is larger than $d(x,z)+d(z,z')+d(z',x')≥d(x,x')$. $\endgroup$ Commented Mar 1, 2020 at 14:55

It seems nobody mentioned spaces of distributions. These are duals of function spaces and they are endowed with the weak-topology. In general this topology is not metrizable. This is a fundamental construction in the modern theory of PDEs and there are plenty of books with many results (Hormander, Gel'fand and Shilov etc.). See also this answer

  • $\begingroup$ One of the other questions remarked that physicists think general topology is uninteresting, since all naturally-occuring spaces are metrizable. But physicists also use distributions widely, without worrying about their foundation. Conclusion: what physicists care about, and what mathematicians think physicists should care about ... these do not coincide. $\endgroup$
    – GEdgar
    Commented Sep 4, 2013 at 13:53

A lot of physicists claim that the abstraction to topological spaces is uninteresting, since all topological spaces that arise naturally in physics are metrizable (homeomorphic to a metric space).

That may be true; however, consider the product space of infinitely many copies of the closed interval $[0,1]$ (the Hilbert cube), which comes up a lot in physics. This is metrizable (indeed, if we consider it as the topologically equivalent $[0,1]\times[0,\frac12]\times[0,\frac13]\times\dots$ then it is seen to be a subspace of the metric space $\ell_2$); however, there is no canonical way of putting a metric on the space that gives you any more interesting information about the space. The only interesting thing about the space is its topology, and its topological properties (note that the same is true in the field of algebraic topology). So it makes no sense to treat it as a metric space.

Also, if you take uncountably many copies of $[0,1]$ then the resulting space is not metrizable, but I don't think it turns up that much in physics either.

  • $\begingroup$ My interest in topological spaces would not be reduced if physicists are not interested. I don't think that physicists are interested in en.wikipedia.org/wiki/Jordan_curve_theorem, however I am interested. Also could you elaborate more on the part " The only interesting about the space is its topology". Finally, thanks for your answer $\endgroup$
    – Amr
    Commented Aug 5, 2013 at 20:01
  • 2
    $\begingroup$ Note that physics relies on integral (e.g. Fourier) transforms, the theory of which relies on the theory of function spaces, the theory of which relies somewhat on general topology. $\endgroup$ Commented Aug 7, 2013 at 3:50
  • 2
    $\begingroup$ Indeed, physicists often deal with much simpler (finite-dimensional!) vector spaces which don't permit any canonical metric to be defined: whenever you consider tuples of quantities with different physical dimension, the only way to define a metric is to make a choice of unit system. But the calculations you do must then not depend on the choice of units, otherwise It'd be unphysical. $\endgroup$ Commented Apr 26, 2015 at 8:35

These are further examples of answers to your third question, related to Funcional Analysis.

Direct Method of the Calculus of Variations: Let $M$ be a weakly closed subset of a reflexive Banach space $E$ and $F\colon M\to\mathbb{R}$ a weakly lower semicontinuous function such that there exists $a\in\mathbb{R}$ for which $F^{-1}(-\infty,a]$ is nonempty and norm-bounded. Then $F$ attains its infimum at some point of $M$.

The hypotheses might look a little complicated, but they are so mostly for the sake of generality.

Intuitively, minimization of functionals should already be interesting. Moreover, these types of problems appear when one is looking for the existence of solutions to PDEs. Sometimes it suffices to minimize some sort of "energy" functional on a Sobolev space (or some space of distributions), which satisfies conditions such as the ones above, in order to conclude that some problem (e.g. a Cauchy problem) has a solution. A very well-known application of this sort is Douglas' solution to the Plateau Problem.

The Direct Method of the Calculus of Variations is a direct application of the Banach-Alaoglu theorem.

Gelfand representation of commutative $C^*$-algebras: Every unital commutative C*-algebra is isomorphic to $C(K)$ for some compact Hausdorff space $K$.

This theorem again requires Banach-Alaoglu and Stone-Weierstrass for non-metrizable spaces.

This is one of the fundamental theorems in the theory of $C^*$-algebras, which allows one to construct a functional calculus and essentially treat self-adjoint elements as if they were functions. You can see more on any book on C*-algebras.

(Although if you only care about separable $C^*$-algebras you could get away with metrizable $K$. I'm not sure if the proof of Gelfand representation could be made significantly easier in this case, though.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .