Prove the closure of E in a Metric Space X is closed. (page 35)(adsbygoogle = window.adsbygoogle || []).push({});

Rudin states:

if p∈X and p∉E then p is neither a point of E nor a limit point of E..

Hence, p has a neighborhood which does not intersect E. (Great)

The compliment ofthe closure of Eis therefore open. WHY? I don't see it...

BTW, I know there are different ways to proving this, but I want to understand the last line jump. Thanks.

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# Rudin's Theorem 2.27

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