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@ -161,7 +161,12 @@ $F(z_k, x_1) \to 0$, $F(z_k, x_2) \to 0$.
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So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
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So $(F_\alpha)_{\alpha \le \beta}$ is a strictly
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increasing chain of closed subsets.
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increasing chain of closed subsets.
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But $X$ is second countable,
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But $X$ is second countable,
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so $\beta$ is countable.
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so $\beta$ is countable:
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Let $\{U_n\} = \cB$ be a countable basis
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and for $\alpha$ let $U_\alpha \in \cB$
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be such that $U_\alpha \cap F_\alpha = \emptyset$
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and $U_\alpha \cap F_{\alpha+1} \neq \emptyset$.
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Then $\alpha \mapsto U_\alpha$ is an injection.
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\end{proof}
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\end{proof}
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