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

## 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 = $\bigcap$N(A,$\epsilon$) for $\epsilon$>0

## 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!

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.