Cardinality of separable metric spaces Show that $\mathbb R$ is cardinally larger than any separable metric space S.
I have been trying to solve this on my own. My idea was to start by mapping the open balls of positive rational radius around the points in the dense set of S to the corresponding in $\mathbb R$. Now any ${x}\subset S$ can be written as a limit of an appropriate sequence of open balls, which would correspond to a similar limit in $\mathbb R$, in case the limit exists.
Then I would need to show that different limits are obtained in $\mathbb R$ for different $x\in S$. This is a step I am not sure how to show. 
Alternatively, one could show that all possible sequences of rationals have no greater cardinality than reals. What would be a good way to show this? I know that all reals can be written as some convergent sequence of rationals, but I am looking for something slightly different here.
I did not find this useful as it does not fill in the details: Every separable metric space has cardinality less than or equal to the cardinality of the continuum.
 A: It is not possible to show that $\mathbb R$ is cardinally larger than any separable metric space; that would mean showing that $\mathbb R$ is larger than itself, since $\mathbb R$ is a separable metric space.
I think you want to show that $\mathbb R$ is cardinally greater than or equal to any separable metric space; or, to put it more naturally, the cardinality of any separable metric space is less than or equal to the cardinality of $\mathbb R.$
Let $S$ be a separable metric space, and let $A$ be a countable dense subset of $S.$ For $x\in S$ and $n\in\mathbb N,$ define
$$f_n(x)=\{a\in A:d(a,x)\lt1/n\}\in\mathcal P(A)$$
and
$$f(x)=(f_1(x),f_2(x),f_3(x),\dots,f_n(x),\dots)\in\mathcal P(A)^\mathbb N.$$
If $x,y\in S,\ x\ne y,$ then taking $n$ sufficiently large we have $f_n(x)\cap f_n(y)=\emptyset.$ Since $f_n(x)\ne\emptyset$ (because $A$ is dense), it follows that $f_n(x)\ne f_n(y)$ and so $f(x)\ne f(y).$ This shows that the map $f:S\to\mathcal P(A)^\mathbb N$ is injective. It follows that
$$|S|\le|\mathcal P(A)^\mathbb N|=|\mathcal P(A)|^{|\mathbb N|}\le(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=|\mathbb R|.$$
Essentially the same argument shows that, more generally, any separable first-countable Hausdorff space has cardinality at most $2^{\aleph_0};$ just define $f_n(x)=A\cap U_n(x)$ where $\{U_n(x):n\in\mathbb N\}$ is a neighborhood base for $x.$
A: Let $X$ be a separable metric space, then there exists a dense numerable subset $A$. 
Suppose that two distinct points $x$ and $y$ have the same distances to every element of $A$. Since $A$ is dense we can pick a point $a\in A$ such that $d(y,a)=d(x,a)< d(x,y)/2$, use the triangle inequality to obtain a contradiction.
Therefore the distances to the elements of $A$ uniquely determine the point of $X$. Clearly the number of such options is $|\mathbb R ^\mathbb N|=|\mathbb R|$.
