Closure of Open Ball is a subset of Closed Ball

Last updated Nov 1, 2022

Let $(X, d)$ be a Metric Space. Let $x \in M$ and let $\epsilon > 0$. Then $\text{cl}B_{\epsilon}(x) \subset \overline{B_{\epsilon}(x)}$.

Recall Closure of a Set in a Metric Space is all its Sequential Limits, so $x \in \text{cl} B(X)$ If and Only If there exists $({x}{n}){n=1}^{\infty} \subset B(X)$ so that $x_{n} \to x’$. Then, because Metrics are Continuous and Components Converge iff Sequence Converges in Finite Product Metric Spaces, $d(x_{n}, x) \to d(x’, x)$. Since $d(x_{n}, x) < \epsilon$ for all $n \in \mathbb{N}$, we have that $d(x’, x) \leq \epsilon$ because of the Order Limit Theorem. Thus $\text{cl} B_\epsilon(x) \subset {x’ \in X : d(x, x’) \leq \epsilon} =: \overline{B_\epsilon(x)}$. $\blacksquare$

1. The Closure of an Open Ball is not always the Closed Ball.

• Closed Ball
• Closure
• Open Ball