Let $X$ be an arbitrary subset of $\mathbb{R}^n$ and, for each $\varepsilon>0$, set $X_{\varepsilon} := \bigcup\limits_{x\epsilon X} \mathbb{B}(x,\varepsilon)$. Prove that $\bigcap\limits_{\varepsilon > 0} X_{\varepsilon} = \bar{X}$.
Where $\mathbb{B}(x,\varepsilon)$ is the open ball centered at $x$ of radius $\varepsilon$ and $\bar{X}$ is the closure of $X$. I'm trying to show that any convergent sequence in $X$ converges to a value in the given set. I'm not sure how to approach the problem. Any help?
Best Answer
If $y \in \overline {X}$ and $\epsilon >0$ the there exists $x \in X$ such that $d(y,x) <\epsilon$ which means $y \in B(x,\epsilon)$. So $\overline {X} \subset \cap_{\epsilon >0} \cup_{x \in X} B(x,\epsilon)=\cap_{\epsilon >0} X_{\epsilon}$.
Conversely suppose $y \in \cap_{\epsilon >0} X_{\epsilon}$. Then for each $n$ there exists $x_n \in X $ such that $B(y,x_n) <\frac 1 n$. Hence $x_n \to y$ which implies $y \in \overline {X}$.