Cardinality of an infinite separable connected metric space is $2^{\aleph_0}$. How to prove:
Cardinality of an infinite separable connected metric space is $2^{\aleph_0}$.
Thanks in advance!!
 A: Let $D$ be a countable dense subset of $X$. For each $x\in X$ let $\mathscr{B}(x)=\{B(x,2^{-n}):n\in\Bbb N\}$, where as usual $B(x,r)=\{y\in X:d(x,y)<r\}$; $\mathscr{B}(x)$ is a local base at $x$. For each $x\in X$ and $n\in\Bbb N$ let $D(x,n)=D\cap B(x,2^{-n})$, and let $\mathscr{D}(x)=\{D(x,n):n\in\Bbb N\}$; $\mathscr{D}(x)$ is a countable family of subsets of $D$. Now show:


*

*$D$ has only $2^{\aleph_0}$ distinct countable families of subsets.

*If $x,y\in X$ and $x\ne y$, then $\mathscr{D}(x)\ne\mathscr{D}(y)$.
This part shows that the cardinality is at most $2^{\aleph_0}$ and doesn’t use connectedness. That comes in when you want to show that the space doesn’t have a smaller cardinality. Remember that $X$ is completely regular, so if you pick a point $x$ and an open set $U$ containing $x$, there is a continuous function $f:X\to[0,1]$ such that $f(x)=1$ and $f(y)=0$ for all $y\in X\setminus U$. If $|X|<2^{\aleph_0}$, that map cannot be a surjection; use that to produce a disconnection of $X$.
A: To see that $|X| \ge 2^{\aleph_0}$, we don't need to use complete regularity.  Pick $x \in X$; I claim that for every sufficiently small $r$, say $r < R$, there exists $x_r \in X$ with $d(x, x_r) = r$.  (Suppose not; then you can show that $B(x,r)$ is open and closed.)  Since the points $\{x_r\}$ are necessarily distinct, you have a set of the cardinality of $(0,R)$.
