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

[missavvy]

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 \epsilon>0, define the sets

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

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

Show that A is closed iff A = \bigcapN(A,\epsilon) for \epsilon>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,\epsilon)?

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!

 

[micromass]

So, for \Rightarrow, you must prove that

A=\bigcap{N(A,\varepsilon)}

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