Every metric space is the union of open balls.

Question

Proof: Let X be a metric space. Note since for x $\in X$ $B(x,r_1) \subseteq B(x,r_2) \subseteq B(x,r_3) \subseteq$ $.....$ if $r_1<r_3<r_4...$, it follows that $\cup_{r \geq 0} B(x,r)$ $=$ X. ( I am not convinced by this argument, can someone make it more clear, please?). Furthermore, if $a \in \cap_{r>0} B(x,r)$ then for every $r > 0$ we have $d(x,a) <r$ and so $a = x$.

Answer 1

Note that\begin{align}X&=\bigcup_{x\in X}\{x\}\\&\subset\bigcup_{x\in X}B(x,1)\\&\subset X\end{align}and therefore $X=\bigcup_{x\in X}B(x,1)$.

Answer 2

Let $y\in X$. If $y\neq x$, then $d(y,x)=\alpha>0$ for some $\alpha>0$ whence $y\in B(x,\alpha+1)$. If $y=x$, then $d(y,x)=0$ whence $y\in B(x,1)$ for example.

The reverse inclusion is obvious.

Answer 3

$B(x, r_1) \subseteq B(x, r_2)$ isn't important for $\cup_{r \geq 0} B(x,r) = X$. If you need just open balls, you can write $X = \bigcup_{x \in X} \{x\} \subseteq \bigcup_{x \in X} B(x, 1)$ as $\{x\} \subseteq B(x, 1)$.

If you want center fixed, you can write $$X = \bigcup_{y \in X} \{y\} \subseteq \bigcup_{y \in X} B(x, d(x, y) + 1) = \bigcup_{r \in \{d(x, y) + 1|y \in X\}} B(x, r) \subseteq \bigcup_{r \geqslant 0}B(x, r)$$.

Answer 4

If we fix $x\in X$ and $y$ is any point in $X$, then $d(x,y)=r$ for some $r \ge 0$ and so $y \in B(x,r+1)$. It follows that $X = \bigcup_{r \ge 0} B(x,r)$, as all balls are subsets of $X$ so the right to left inclusion is obvious while I just showed the reverse implication that any $y \in X$ is in one of these balls.

That $B(x,r_1) \subseteq B(x,r_2)$ for $r_1 \le r_2$ is obvious from the definition of balls: $y \in B(x,r_1) \iff d(x,y) < r_1 \,(\le r_2) \implies d(x,y) < r_2 \iff y \in B(x,r_2)$ so $X$ is indeed a union of an increasing family of balls.

We can even take all $r$ to be in $\mathbb N$ as for each $r \in \mathbb{R}$, there is some $n\in \mathbb N$ with $r \le n$ so we can make it a countable union if so desired. So $$X=\bigcup\{B(x,n): n \in \mathbb{N}\}$$ for any $x \in X$.