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?
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 $|X|\le2^{|\mathcal B|}.$