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I was wondering right that since the notion of a topology is much more general than that of a metric, and that "neighborhodness", if you will, and the concept of continuity, is generalized by the notion of a topology. so is the set of all topological spaces actually bigger than that of metric spaces? in other words if $$\mathscr{T}=\{x|x \text{ is a topological space}\}$$ and if $$\mathscr{M}=\{x|x\text{ is a metric space}\}$$ then is $$\text{card}\mathscr{T}>\text{card}\mathscr{M}$$ or are they equal? in both cases how can we prove that?

or perhaps the sets $\mathscr{T}$ and $\mathscr{M}$ don't even exist at all similar to how the set of all sets doesn't exist?

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  • $\begingroup$ i know that a metric space is also a topological space, and that a topology induces a metric, however even then it may be that the sets are equal just llike $\text{card}[0,1]=\text{card}\mathbb{R}$ $\endgroup$
    – user153330
    Feb 4, 2016 at 23:23
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    $\begingroup$ In the sense of ZFC, there is no set of all topological spaces: any set can be given the discrete topology, and the set of all sets is "too large to be a set". Similarly, any set can be given the discrete metric, which creates the same problem. A more reasonable question would be to make this comparison with a fixed bound on the maximum cardinality of the underlying set of the spaces in question. (For instance you could take it to be $\beth_1$, i.e. $|\mathbb{R}|$.) $\endgroup$
    – Ian
    Feb 4, 2016 at 23:26
  • $\begingroup$ @Ian so what would be the answer then if we took it to be $\text{card}\mathbb{R}$? $\endgroup$
    – user153330
    Feb 4, 2016 at 23:41
  • $\begingroup$ $\mathscr{T}$ and $\mathscr{M}$ are proper classes, not sets. In fact they can be put in 1-1 correspondence. But there are topologies which are not induced by metrics — that is, there are spaces that are not metrizable. $\endgroup$
    – BrianO
    Feb 5, 2016 at 0:03
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    $\begingroup$ @BrianO: I'm intrigued to know how you would put $\cal T$ and $\cal M$ in 1-1 correspondence. But more interestingly, the existence of non-metrizable spaces doesn't obviously imply anything about the number of topologies on a set of a given cardinality and the number of metrics on that set. $\endgroup$
    – Rob Arthan
    Feb 5, 2016 at 0:09

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As others have mentioned, the collection of all metric spaces and the collection of all topological spaces both form proper classes, so we can't reason about their size using cardinality.

A similar question that can be answered is: given a set $X$, are there more topologies on $X$ than metrics on $X$?

To this end, given a set $X$, let $T(X)$ be the set of topologies on $X$ and let $M(X)$ be the set of metrics on $X$. (For simplicity, let's say $X$ is infinite.) Then $M(X) \subseteq \mathbb{R}^{X \times X}$; and $|\mathbb{R}^{X \times X}| = 2^{|X|}$, so there are at most $2^{|X|}$ metrics on $X$; but (as in this answer) there are $2^{2^{|X|}}$ topologies on $X$.

Since $2^{|X|} < 2^{2^{|X|}}$ for all sets $X$, it follows that on any infinite set there are strictly more topologies than metrics.

Interestingly, exactly the opposite is true for finite sets: if $X$ is finite then there are uncountably many metrics on $X$ but only finitely many topologies on $X$.

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    $\begingroup$ What about the number of topologies v. metrics up to homeomorphism or isomorphism? I think the question you refer to implies that the result is the same. Is that correct? $\endgroup$
    – Rob Arthan
    Feb 5, 2016 at 0:19
  • $\begingroup$ @RobArthan: For topologies there is indeed the same number up to homeomorphism (as long as $X$ is infinite), by the linked answer. For metric spaces up to homeomorphism, I don't know the answer in general, but see this answer as long as $|X|\leq \aleph_1$ or $|X|\geq 2^{\aleph_0}$. $\endgroup$
    – Eric Wofsey
    Feb 5, 2016 at 1:28
  • $\begingroup$ For metrics up to isomorphism, you can encode a graph on $X$ in a metric by letting $d(x,y)=1$ if there is an edge between $x$ and $y$ and $d(x,y)=2$ if there is not (for $x\neq y$), and use the fact that there are $2^{|X|}$ non-isomorphic graphs on $X$ (e.g., by encoding total orders in a graph by a method similar to how I encoded total orders in a metric in my answer linked above). $\endgroup$
    – Eric Wofsey
    Feb 5, 2016 at 1:32
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$\mathscr{T}$ and $\mathscr{M}$ are proper classes in ZFC, not sets. (They're also proper classes in set theories that can talk about proper classes explicitly — von Neumann-Godel-Bernays set theory and Morse-Kelley set theory.)

In fact the two classes can be put in 1-1 correspondence: see the answer by Q the Platypus, which points out that the Schroeder-Bernstein construction applied to injections $V \leftrightarrows \mathscr{M}$ will work to yield a bijection $V \xrightarrow{\sim} \mathscr{M}$, and similarly for $\mathscr{T}$.

Neither of these classes can be bijected with a cardinal: a cardinal is a set, and by the Axiom of Replacement, the image of a set under a function is also a set. So $card(\mathscr{T})$ and $card(\mathscr{M})$ are simply undefined.

There are topologies which are not induced by metrics — that is, there are spaces that are not metrizable. In that sense, $\mathscr{T}$ is larger than $\mathscr{M}$ — more precisely, $\mathscr{T}$ properly includes the class of topologies induced by members of $\mathscr{M}$.

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    $\begingroup$ Global choice isn't required--as indicated in Q the Platypus's answer, it is easy to define injections from $V$ to each of $\mathscr{T}$ and $\mathscr{M}$, and then by Schroder-Bernstein you can define bijections between both of them and $V$. $\endgroup$
    – Eric Wofsey
    Feb 5, 2016 at 1:15
  • $\begingroup$ @EricWofsey True that. I emended accordingly. $\endgroup$
    – BrianO
    Feb 5, 2016 at 1:51
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The class of all sets is bijective to the class of all metric spaces. Because every set can be injectively mapped to a metric space via the trivial metric and since every metric space is a pair of sets there is a injective map from the the class of all metric spaces back to the class of all sets.

Likewise since the class of all topological spaces is a subclass of the class of all sets there is a bijection between it and the class of all metric spaces.

This answer depends on Bernstein-Schröder theorem holding for proper classes. This is unprovable in ZFC because ZFC has no concept of a proper class (or bijections between classes). However this is true in systems that have definitions for proper classes and mappings between them.

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    $\begingroup$ That is actually incorrect. There is a Cantor-Bernstein theorem for classes, even in ZF. Given two formulas which define classes, and there are two formulas which define injections between those classes, there is a formula which define a bijection between the two classes. $\endgroup$
    – Asaf Karagila ♦
    Feb 5, 2016 at 6:47
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Here's an answer for the case of finite spaces: there are countably many finite topological spaces up to homeomorphism (because each finite set admits only a finite number of topologies). There are uncountably many finite metric spaces up to isometry (because just for a space with two points $x$ and $y$ you have uncountably many choices for $d(x, y)$).

This complements Clive Newstead's answer: if you look at a particular $X$: if $X$ is finite it has more metrics than topologies, if it is infinite it has more topologies than metrics.

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The other answers here deal (correctly) with cardinality questions about the classes you're asking about. I don't think that's the way to ask the question that seems to interest you. Perhaps you want to know in what sense there are more topological spaces than metric spaces. If that's your question then ...

Every metric space is a topological space in a natural way - a metric determines a topology. (Different metrics on the same space may or may not determine the same topology.) But there are topological spaces where the topology does not come from any metric. See https://en.wikipedia.org/wiki/Metrization_theorem and https://mathoverflow.net/questions/52032/examples-of-non-metrizable-spaces for examples.

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