show that $F= \underset{n\geqslant1}{\bigcap} \left\{ x \in X, d(x,F) < \frac{1}{n} \right\}.$ Let $F$ be a closed set from a metric space $(X,d)$ , show that $$F= \underset{n\geqslant1}{\bigcap} \left\{ x \in X, d(x,F) &lt; \frac{1}{n} \right\}.$$
My attempt :  $\Rightarrow)$
for all $x \in X$,  $ x\in F\iff d(x,F)=0$, then  $x \in \underset{n\geqslant1}{\bigcap} \left\{ x \in X, d(x,F) &lt; \frac{1}{n} \right\} $
$\Leftarrow)$ for all $n\geqslant 1$, $0\leqslant d(x,F) &lt; \frac{1}{n} \xrightarrow[n\to \infty]{}0,$ then $0\leqslant d(x,F) &lt;0 $ !! ( can a strict inequality became broad ? this is my question ? )
Generally, limits always transform strict into large hence the result.
(Strict always implies large actually)
Think of $a_n=\frac{1}{n}$ :
$$0&lt;a_n$$ implies $$0\leq a_n$$ which is convenient with taking the limit
Generally, limits always transform strict into large hence the result.
(Strict always implies large actually)
Think of $a_n=\frac{1}{n}$ :
$$0&lt;a_n$$ implies $$0\leq a_n$$ which is convenient with taking the limit