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A set is closed iff it equals an intersection of closed sets

  • Thread starter missavvy
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Homework Statement



Let M be a metric space, A a subset of M, x a point in M.

Define the metric of x to A by

d(x,A) = inf d(x,y), y in A

For [itex]\epsilon[/itex]>0, define the sets

D(A,[itex]\epsilon[/itex]) = {x in M : d(x,A)<[itex]\epsilon[/itex]}

N(A,[itex]\epsilon[/itex]) = {x in M: d(x,A)[itex]\leq[/itex][itex]\epsilon[/itex]}

Show that A is closed iff A = [itex]\bigcap[/itex]N(A,[itex]\epsilon[/itex]) for [itex]\epsilon[/itex]>0

Homework Equations





The Attempt at a Solution



I was able to do the <= implication.
I'm having trouble doing the => one...

If I assume A is closed, all I know is that its compliment is open, which is a union of open sets... then should I just use that fact and DeMorgans laws to show A is in fact an intersection of those closed N(A,[itex]\epsilon[/itex])?

OR should I take an arbitrary point in A, and show it belongs to the intersection..??

Just need some help on what method I should try to use.

Thanks!
 

Answers and Replies

  • #2
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So, for [itex]\Rightarrow[/itex], you must prove that

[tex]A=\bigcap{N(A,\varepsilon)}[/tex]

There are 2 things to prove now: [itex]\subseteq[/itex] and [itex]\supseteq[/itex]. Only [itex]\supseteq[/itex] is nontrivial. For this: take a point in the intersection, and prove that it is in A.
 

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