Prove that $\mathbb B (x_0,r)$ is open and $\overline{\mathbb B} (x_0,r)$ is closed in $X$

Question

Let $(X,d)$ be a metric space. For $x_0 \in X$ and $r>0$, we define the balls $\mathbb B (x_0,r)$ and $\overline{\mathbb B} (x_0,r)$ by $$\mathbb B (x_0,r) = \{x \in X \mid d(x,x_0) < r\}, \quad \overline{\mathbb B} (x_0,r) = \{x \in X \mid d(x,x_0) \le r\}$$

Prove that $\mathbb B (x_0,r)$ is open and $\overline{\mathbb B} (x_0,r)$ is closed in $X$.

Could you please verify if my proof looks fine or contains logical gaps/errors? Any suggestion is greatly appreciated.

My attempt:

1. $\mathbb B (x_0,r)$ is open in $X$

For $x \in \mathbb{B}(x_0, r),$ set $s =d(x_0, x)$ and $\varepsilon =r-s > 0$. For all $y \in \mathbb{B} (x, \varepsilon)$, we have $$d(x_0, y) \le d(x_0,x)+d(x,y) < s + \varepsilon = r$$ and so $\mathbb{B}(x, \varepsilon) \subseteq \mathbb{B}(x_0, r)$. This shows that $x$ is an interior point of $\mathbb{B}(x_0, r)$. Hence $\mathbb B (x_0,r)$ is open in $X$.

2. $\overline{\mathbb B} (x_0,r)$ is closed in $X$

Let $(x_k)$ be a sequence in $\overline{\mathbb B} (x_0,r)$ such that $\lim_{k \to \infty} x_k = x \in X$. Assume the contrary that $x \notin \overline{\mathbb B} (x_0,r)$. Then $d(x,x_0) > r$. Let $s = d(x,x_0) - r > 0$. Since $\lim_{k \to \infty} x_k = x$. There exists $N \in \mathbb N$ such that $d(x_k,x) < s$ for all $k \ge N$. As such, $$\begin{aligned}d(x_N,x_0) &\ge |d(x,x_0) - d(x_N,x)| &&= |(s+r) -d(x_N,x)|\\ &= |r + (s-d(x_N,x))| &&= r + (s-d(x_N,x))\\ &> r\end{aligned}$$ Hence $x_N \notin \overline{\mathbb B} (x_0,r)$, which is a contradiction. As such, $\overline{\mathbb B} (x_0,r)$ is closed in $X$.

Answer 1

$According \ to \ me$, the both of the answers are fine.

An alternative approach towards the second question:

Let us denote the derived set of $\overline{B}_r(x_0)$ by $D_B$. Suppose, $\exists \ w \in D_B$ such that $w \notin \overline{B}_r(x_0)$. Then, by the definition of limit point, $\forall \epsilon>0, \ \exists \ \alpha \in \overline{B}_r(x_0) \text{ such that } d(w,\alpha)<\epsilon$.

Now, $d(w,x_0)>r \implies \epsilon+r> d(w,\alpha)+d(\alpha,x_0) \geq d(w,x_0)>r$.

$\epsilon>0$ is arbitrary, so $r>d(w,x_0)>r$, absurdity.