Closure of Open Unit Ball is Closed Unit Ball
# Statement
Let $X$ be a Normed Vector Space. Then $\text{cl}B(X) = \overline{B(X)}$.
# Proof
Closure of Open Ball is a subset of Closed Ball so $\text{cl}B(X) \subset \overline{B(X)}$. To see the other direction, note that for $x \in X$ so that $||x|| = 1$, we can construct a Sequence $x_{n} \to x$ from $B(X)$ by picking $x_{n} = (1-\frac{1}{n})x$. Since $||x_{n}|| = (1-\frac{1}{n})||x|| = \left(1-\frac{1}{n}\right)< 1$, $x_{n} \in B(X)$ as we desired. Then, $||x_{n} - x|| = \frac{1}{n}||x|| = \frac{1}{n} \to 0$, so $x_{n} \to x$. Because Closure of a Set in a Metric Space is all its Sequential Limits, $x \in \text{cl}B(X)$. $\blacksquare$