Is the cardinality of the topology of a separable infinite metric space uncountable

Question

Given a metric space $X$ which is separable and infinite, can we say that $\tau$ (its topology) has cardinality greater than that of the natural numbers? And if so how to go about it.

Answer 1

HINT: If $X$ is discrete, then $\tau=\wp(X)$, so $|\tau|\ge|\wp(\Bbb N)|=2^{\aleph_0}>\aleph_0=|\Bbb N|$.

If not, then there is an $x\in X$ such that $x\in\operatorname{cl}(X\setminus\{x\})$.

• Explain why there is a sequence $\langle x_n:n\in\Bbb N\rangle$ of distinct points converging to $x$.

Let $D=\{x_n:n\in\Bbb N\}$.

• Show that $D$ is discrete: for each $n\in\Bbb N$ there is an open nbhd $U_n$ of $x_n$ such that $U_n\cap D=\{x_n\}$.
• Show that if $A,B\subseteq\Bbb N$, and $A\ne B$, then $\bigcup_{n\in A}U_n\ne\bigcup_{n\in B}U_n$.
• Conclude that the map $$\varphi:\wp(\Bbb N)\to\tau:A\mapsto\bigcup_{n\in A}U_n$$ is an injection, and hence $|\tau|\ge|\wp(\Bbb N)|=2^{\aleph_0}>\aleph_0=|\Bbb N|$.