Metrizable implies first-countable

Let $(𝑋, 𝑑)$ be a metric space. Then $𝑋$ is first-countable under its metric topology. #m/thm/topology

Proof

The following set defines a (nested) neighbourhood basis for a point $𝑥 \in 𝑋$ :
$B (𝑥) = {B 1 / 𝑘 (𝑥)} \infty 𝑘 = 1$
For any open neighbourhood $𝑈$ of $𝑥$ must contain an open ball $B 𝛿 (𝑥)$ for $0 < 𝛿 \leq 1$ . Letting $𝑘 = ⌈ 𝛿 - 1 ⌉$ , it follows the basic open neighbourhood $B 1 / 𝑘 (𝑥) \subseteq B 𝛿 (𝑥) \subseteq 𝑈$ .

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Metrizable implies first-countable

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