Prove that $\cup_{n=0}^{\infty}B_n$, where $B_0=\{x\}$ and $B_{n+1}=\{z\in X\vert \exists b\in B_n s.t. d(z,b)<\epsilon\}$, is a clopen set

Question

Let $(X,d)$ be a metric space. A chain in $X$ from $p$ to $q$ is a finite sequence of points in $X$, $\{x_0,x_1,...,x_n\}$ s.t. $x_0=p$, $x_n=q$ and $d(x_i,x_{i+1})\leq \epsilon$ for $i=0,1,...,n-1$. $X$ is called $\epsilon$ chainable if there exists an $\epsilon$ chain for any two points $p,q\in X$.

We need to prove that if $X$ is connected then it is $\epsilon$-chainable for any $\epsilon >0.$ For this purpose, we define a sequence of subsets of $X$ as shown: let $B_0=\{x\}$ and $B_{n+1}=\{z\in X\vert \exists b\in B_n s.t. d(z,b)<\epsilon\}$. To show that $X$ is $\epsilon$-chainable for any $\epsilon >0$ we need to show that $\cup_{n=0}^{\infty}B_n$ is a clopen set.

My attempt: $\cup_{n=0}^{\infty}B_n$ is open as the union of open sets and $B_0\subset B_1$. How to prove that $\cup_{n=0}^{\infty}B_n$ is closed?

Answer 1

HINT

Let $y$ be a limit point of the union. Consider an $\epsilon $ ball around $y$. It contains a point from the union and hence some $B_n$. So $y$ should be in $B_{n+1}$

Answer 2

You could use the chain lemma (which is close in spirit to this idea); I show and formulate it here:

$X$ is connected iff for every open cover $\mathcal{U}$ of $X$, for every two $x,y \in X$ there is a chain from $\mathcal{U}$ from $x$ to $y$, i.e. $U_0, U_1, \ldots U_n \in \mathcal{U}$ such that $x \in U_0, y \in U_n$ and for all $i=0,\ldots n-1$: $U_i \cap U_{i+1} \neq \emptyset$.

Having this useful tool makes it quite easy: using the left to right implication on the open cover $B(x, \frac{\varepsilon}{2})$ and use the centres of the balls from the chain as your $\varepsilon$-chain: for intersecting balls the radii are $< \varepsilon$ apart.

But for metric spaces in general, we cannot show connectedness from the presence of such chains (e.g. $(0,1) \cup (1,2)$ is disconnected but any two points are connected by an $\varepsilon$-chain for any $\varepsilon >0$), but for compact metric spaces we can (if such a space is disconnected, the decomposing sets are compact as well, and thus some distance apart, and take half that distance as the $\varepsilon$ etc.). The cover condition is more general, and also is used to see that a locally path-connected connected $X$ is path-connected and more facts along this line.