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A regular space $X$ is a topological space for which one point sets are closed and for every $x\in X$ and for every closed set $C\subset X\backslash \{x\}$, there exist disjoint open sets containing $x$ and $C$, respectively.

Assume $X$ is regular and has a countable basis. In the proof of the Urysohn Metrization Theorem, it is shown that $X$ can thus be imbedded in $\mathbb{R}^\omega$. Since imbeddings are injective, an interesting consequence of this is that $\text{card}(X)\leq 2^{\aleph_0}$.

But this fact never would have been intuitive to me without the metrization theorem, which required Urysohn's lemma to prove. Does anyone know of a quicker way to see why this must be true? Also, can the conditions be relaxed in any way to obtain the same result?

confused_wallet
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    @bof thanks, I edited the question to change $\aleph_1$ to $2^{\aleph_0}$. To your second comment: in the definition of regular, I am assuming 1-point sets to be closed. – confused_wallet Feb 12 '18 at 00:11

1 Answers1

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In fact $2^{\aleph_0}$ is the maximum cardinality for a T$_0$ space with a countable base. Suppose $X$ is a T$_0$ space and $\mathcal B$ is a base for the topology of $X.$ Then the map $x\mapsto\{B\in\mathcal B:x\in B\}$ is an injection, showing that $|X|\le2^{|\mathcal B|}.$

bof
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