Closure and boundary of an open ball Let $B=\{x\in\mathbb{R}^n|\Vert x\Vert_2<1\}$. How can I show rigorously the boundary of it is $S=\partial B=\{x\in\mathbb{R}^n|\Vert x\Vert_2=1\}$ and the closure is $D=\text{cl} (B)=\{x\in\mathbb{R}^n|\Vert x\Vert_2\leqslant1\}$?
My idea is to consider the norm function $h:\mathbb{R}^n\to\mathbb{R}$ which is continuous: $\text{cl}(h^{-1}[(-\infty,1)])\subseteq h^{-1}[\text{cl}((-\infty,1))]=h^{-1}[(-\infty,1]]$. But the other direction is not true in general. So how can we show it?
 A: Suppose that $\|z\|\leqslant1$ (since there's only one norm here, I will use $\|\cdot\|$ instead of $\|\cdot\|_2$). Given $\varepsilon>0$, let $y=\left(1-\frac\varepsilon2\right)z$. Then $\|y\|=1-\frac\varepsilon2<1$ and $\|z-y\|=\frac\varepsilon2<\varepsilon$. This proves that$$z\in\overline{\{x\in\mathbb{R}^n\,|\,\|x\|<1\}}.$$
On the other hand, if $\|z\|>1$, then it follows from the triangle inequality that$$\{y\in\mathbb{R}^n\,|\,\|y-z\|<\|z\|-1\}\cap\{x\in\mathbb{R}^n\,|\,\|x\|<1\}=\emptyset.$$Therefore, $z\notin\overline{\{x\in\mathbb{R}^n\,|\,\|x\|<1\}}$. So, I've proved that$$\overline{\{x\in\mathbb{R}^n\,|\,\|x\|<1\}}=\{x\in\mathbb{R}^n\,|\,\|x\|\leqslant1\}.$$
Since $\{x\in\mathbb{R}^n\,|\,\|x\|<1\}$ is an open set, it is equal to its interior. And since the boundary of $\{x\in\mathbb{R}^n\,|\,\|x\|<1\}$ is its closure minus its interior, the boundary is equal to $\{x\in\mathbb{R}^n\,|\,\|x\|=1\}$.
A: Let $x\in S$. Then $\|x\|_2 = 1$. Define a sequence $\{x_n\}_{n\in\mathbb{N}}$ in the following way,
\begin{align}
x_n = \left(1-\frac{1}{n+1}\right)x
\end{align}
Then we have that $\|x_n\|_2 = \|\left(1-\frac{1}{n+1}\right)x\|_2=\left(1-\frac{1}{n+1}\right)\implies x_n\in B$. So the sequence $\{x_n\}_{n\in\mathbb{N}}$ is contained in $B$, if we show its limit point is $x$ then we are done, since then we could conclude that $x$ is in the boundary and since $x$ was arbitrary in $S$ we could conclude that $S = \partial B$.
Let $\epsilon > 0$ and choose $N$ such that $N>\frac{1}{\epsilon}-1$, then for $n\geq N$ we have,
$$\|x_n-x\|_2 = \|\left(1-\frac{1}{n+1}\right)x-x\|_2 = \frac{1}{n+1} < \epsilon$$
So the sequence converges to $x$ and we are done showing that $\partial B =S$.
The closure of a set can be written, $\bar{C} = C\cup\partial C$. It is obvious that $S\cup B=\bar{B}$.
