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Bachelier
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Prove the closure of E in a Metric Space X is closed. (page 35)
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 of the closure of E is 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.
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 of the closure of E is 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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