# Thread: a closed set of X is the intersection of countably many open subsets of X

1. ## a closed set of X is the intersection of countably many open subsets of X

My question is only concerning part (vi) of the question.
Sorry for posting the whole question, but I thought it might help someone to understand the context in which the question arises.
I got up to part (vi). I was wondering if the right track was to define $E_n:=\{x;d_E(x) < \frac{1}{n}\}
$
, take the intersection of all the $E_n$, and prove that this new set is closed. If yes, I did not succeed, a little hint would be appreciated.
Let X be a metric space and let E be a nonempty subset of X. ( $E \subset X
$
) Define $d_E(x) = inf\{d(x,e);e \in E\}
$
.

(i) Use the triangle inequality to show that $d_E(x) \leq d(x,x') + d_E(x')
$
.
(ii) Switching the roles of x and x′ and combining, show that $\left| d_E(x)-d_E(x') \right| \leq d(x,x')
$
.
(iii) Deduce that $d_E : X \rightarrow [0, \infty[
$
is a continuous function.
(iv) If in addition E is a closed subset of X, show that $d_E(x)=0 \iff x \in E
$
.
(v) Deduce from (iii) that for fixed t > 0, $\{x;d_E(x) < t\}
$
is an open subset of X containing E.
(vi) Choosing $t = \frac{1}{n}
$
, $n \in \mathbb{N}
$
in (v) show that every closed set of X can be written as the intersection of countably many open subsets of X.

2. I have a hard time understanding the notation.
That said, I take $dE(x)$ to be the distance from the point $x$ to the set $E$.

It is worth noting that if a set $M$ is closed if and only if $M=\{x:dM(x)=0\}$.

In this case the sets $E_n$ are a countable collection of open sets.

So yes, you are on the right tract.

3. I tried to make the notation more clear using latex.
Can't you have an open subset M of X such that $
M=\{x:d_M(x)=0\}$
(Shouldn't it happen for any subset in fact?). Or is it that the boundary points are included in the set with the definition of M?

4. Originally Posted by vincisonfire
I tried to make the notation more clear using latex.
Can't you have an open subset M of X such that $
M=\{x:d_M(x)=0\}$
(Shouldn't it happen for any subset in fact?). Or is it that the boundary points are included in the set with the definition of M?
An open set contains none of its boundary points.
Boundary points are a distance of zero from the set.
(There are sets that are both open and closed.)

Here are more hints.
$\forall n\left[ {E \subseteq E_n } \right]$.

If $E$ is closed then $t \notin {\rm E}\quad \Rightarrow \quad D_E (t) > 0$.
So $\left( {\exists J} \right)\left[ {\frac{1}{J} < D_E (t)} \right]\quad \Rightarrow \quad t \notin {\rm E}_J$