Cardinality of separable metric spaces

Solution 1:

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 $|\mathbb R ^\mathbb N|=|\mathbb R|$.

Solution 2:

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 $x.$