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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!