How many metrics are there on a set up to topological equivalence? I want to find the number of topologically nonequivalent metrics on a set.
I think if the cardinal of set is finite then we have one metric that is the discrete metric  and every metric on this set is equivalent with the discrete metric. Now what is the number of nonequivalent metrics on an infinite  set? 
 A: Let $S$ be an infinite set.  A metric on $S$ is a function $S\times S\to \mathbb{R}$, and there are only $$|\mathbb{R}|^{|S\times S|}=(2^{\aleph_0})^{|S|}=2^{\aleph_0\cdot |S|}=2^{|S|}$$
such functions.  Thus a trivial upper bound on the number of metrics on $S$ up to equivalence is $2^{|S|}$.  (In contrast, there are $2^{2^{|S|}}$ different topologies up to homeomorphism on $S$, so the restriction to metric spaces in this question really does make a difference!)
There are actually two reasonable definitions of when two metrics on $S$ are "topologically equivalent".  The first definition is that they give the same topology, i.e. that the identity map $S\to S$ is a homeomorphism from the first metric to the second.  With this definition, it is easy to see that there are always $2^{|S|}$ inequivalent metrics on $S$.  For instance, for any subset $T\subseteq S$, you can put a metric on $S\times (\mathbb{N}\cup\{\infty\})$ such that with respect to that metric, the sequence $(s,n)$ converges to $(s,\infty)$ iff $s\in T$.  Choosing a bijection between $S$ and $S\times (\mathbb{N}\cup\{\infty\})$, this gives inequivalent metrics on $S$ for each subset $T\subseteq S$.
The second and more interesting definition of "topologically equivalent" metrics on $S$ is that the two metrics give rise to homeomorphic metric spaces (but the homeomorphism need not be the identity map).  That is, we are asking how many metrizable topological spaces of a given cardinality there are up to homeomorphism.  With this definition, it is also true that the are $2^{|S|}$ inequivalent metrics on $S$, but the proof is more difficult.
Before diving into the proof, let us review the notion of Cantor-Bendixson rank and make some related definitions.  If $X$ is a topological space, write $D(X)$ for the set of non-isolated points of $X$ (the "Cantor-Bendixson derivative of $X$").  For each ordinal $\alpha$, define $D^\alpha(X)$ by induction:


*

*$D^0(X)=X$

*$D^{\alpha+1}(X)=D(D^\alpha(X))$

*For $\alpha$ limit, $D^\alpha(X)=\bigcap_{\beta<\alpha} D^\beta(X)$.
For $x\in X$, the least $\alpha$ such that $x\not\in D^{\alpha+1}(X)$ is called the Cantor-Bendixson rank of $x$ (if no such $\alpha$ exists, we will say $x$ is unranked).  If $x\in X$ is unranked say that the type of $x$ is the lim sup of the ranks of ranked points approaching $x$ plus 1 (i.e., $\inf_U\sup_{y\in U\text{ ranked}}(\operatorname{rank}(y)+1)$, where $U$ ranges over all neighborhoods of $x$).  Equivalently, the type of $x$ is what the rank of $x$ would be if you removed all other unranked points from the space $X$.
Lemma 1: For each ordinal $\alpha>0$, there exists a metrizable space $X_\alpha$ such that $|X_\alpha|=|\alpha|+\aleph_0$, every point of $X_\alpha$ has rank $\leq\alpha$, and there exists a point $x_\alpha\in X_\alpha$ of rank $\alpha$.
Proof: Use induction on $\alpha$.  Assuming you have such spaces $X_\beta$ for each $\beta<\alpha$ and points $x_\beta\in X_\beta$ of rank $\beta$, let $X_\alpha$ be a disjoint union of countably many copies of each $X_\beta$ and one additional point $x_\alpha$, with the distance from $x_\alpha$ to the $n$th copy of $x_\beta$ being $1/n$.  (For $\beta=0$, take $X_\beta$ to be a point.)
Lemma 2: For each ordinal $\alpha>0$, there exists a metrizable space $Y_\alpha$ with $|Y_\alpha|=|\alpha|+\aleph_0$ which contains an unranked point of type $\alpha$ and no unranked points of type $\beta$ for any nonzero $\beta\neq\alpha$.
Proof: Let $Y_\alpha=X_\alpha\sqcup\mathbb{Q}/\sim$, where $\sim$ identifies $x_\alpha\in X_\alpha$ with $0\in\mathbb{Q}$.
We can now construct $2^{|S|}$ nonhomeomorphic metrizable spaces of cardinality $|S|$ for any infinite set $S$.  Namely, if $A$ is any set of $|S|$ nonzero ordinals, all of which have cardinality $\leq |S|$, we can take the space $Z_A=\coprod_{\alpha\in A} Y_\alpha$.  We can recover $A$ from $Z_A$ as the set of nonzero ordinals $\alpha$ such that $Z_A$ contains an unranked point of type $\alpha$.  Every such $Z_A$ has cardinality $|S|$, and $Z_A$ is homeomorphic to $Z_B$ iff $A=B$.  Since there are $2^{|S|}$ such sets $A$, this gives $2^{|S|}$ nonhomeomorphic metrizable spaces of cardinality $|S|$.
If $|S|\geq 2^{\aleph_0}$, there is another rather different way to get $2^{|S|}$ metrizable spaces of cardinality $|S|$.  Given any total order $\leq$ on $S$, we can form a metric space which "draws a picture of the order" as follows.  For each point of $S$, we have a 2-dimensional disk in the metric space.  Whenever $x<y$, draw a path from the center of the disk corresponding to $x$ and the center of the disk corresponding to $y$, and give the path a bit of "ornamentation" that allows you to tell which direction it is going (for instance, you could attach a $3$-dimensional disk somewhere along the path close to $x$, and a $4$-dimensional disk somewhere along the path close to $y$).  Define a metric on the space you obtain from this in the obvious way (the distance between two points is the length of the shortest path you can draw between them, with the obvious interpretation of "length").  It is then straightforward to see that you can recover the ordered set $(S,\leq)$ up to isomorphism from this metric space up to homeomorphism.  Furthermore, the cardinality of the metric space is $|S|\cdot 2^{\aleph_0}=|S|$.  Since there are $2^{|S|}$ non-isomorphic total orders on $S$, this gives $2^{|S|}$ non-homeomorphic metric spaces of cardinality $|S|$.
