Is the set of all topological spaces bigger than the set of all metric space? 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?
 A: 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.
A: $\mathscr{T}$ 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 $\mathscr{M}$.
A: 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.
A: 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.
A: 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 $X$.
