Second countable implies Lindelöf

Let $𝑋$ be a topological space. If $𝑋$ is second-countable then it is Lindelöf.

Proof

Let ${𝑆 𝑛} 𝑛 \in ℕ$ be a countable topological basis of $𝑋$ , and ${𝑈 𝛼} 𝛼 \in 𝐼$ be an open cover. Let $𝐽 \subseteq ℕ$ such that

𝑛 \in 𝐽 ⟺ \exists 𝛼 \in 𝐼 : 𝑆 𝑛 \subseteq 𝑈 𝛼

And for every $𝑛 \in 𝐽$ let $𝛼 𝑛 \in 𝐼$ such that $𝑆 𝑛 \subseteq 𝑈 𝛼 𝑛$ . Since every $𝑈 𝛼$ is the union of some family of $𝑆 𝑛$ with $𝑛 \in 𝐽$ , ${𝑆 𝑛} 𝑛 \in 𝐽$ is a countable open cover of $𝑋$ and therefore ${𝑈 𝛼 𝑛} 𝑛 \in 𝐽$ is too.

The proof relies on the Axiom of Choice.

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