The Closure of a Set is Closed || Metric Spaces Proof

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Let X be a metric space and Y a subset of X. In this video I prove that the closure of Y is closed.

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thank you very much, you explained very well!

yonatankarni

I'll finish watching this later, but right off the top, this is what I didn't get about real analysis. I'll also guess that you'll do a far better job of defining the terms than my class did. thx for your tremendous efforts into this subject.

SequinBrain

I'm trying to learn some stuff. It helped me. Thank you.

ChristopherEvenstar

Pretty much follows immediately from the definition. The closure is the smallest closed set containing that set, and since the closure of Y contains itself, and is also closed, therefore the closure of the closure of Y is just the closure of Y.

persistenthomology

Excuse me but shouldnt it be like this:the intersection of punctured neighbprhood of X and Y must be nonempty?

mzarinchang

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