I found this question in a text book and wanted to ask for some help. I am not quite sure where to start with this.
Prove that in a normed space
$\overline{B^o(x, r)} = B(x, r) \ \ \forall x ∈ X \text{ and } r > 0$.
Hint 1: You need to show that every point $y ∈ X$ such that $||y − x|| = r$ is a limit point of $B^o(x, r)$. To do so, use the sequence $y_n = x + (y − x)(1 − \frac{1}{n})$ and show that $y_n ∈ B^o(x, r)$ $\forall n \text{ and }y_n → y$.
Hint 2: You also need to show that every point $y ∈ X$ such that $||y − x|| > r$ is not a limit point. To do so, use the observation that $X\setminus B(x, r)$ is open.
I guess I can see how to start with Hint 1. $\vert \vert y_n-x \vert \vert=\vert \vert (y-x)\frac{1}{n} \vert \vert \to 0$ ... but then Im not sure
Hint 2 I cant approach as I do not fully understand Hint 1.
When using Hint 1, you made an error in "$\|y_n-x\|=\|(y-x)\frac{1}{n}\|\to0$". Note that $$y_n-x=\left[x +(y−x)\left(1−\frac{1}{n}\right)\right]-x=(y−x)\left(1−\frac{1}{n}\right),$$ which is not what you said. Instead, from the correct expression above, we can find the norm of $y_n-x$. You should work it out and see that this norm is less than $r$, which proves the first part of Hint 1, that all $y_n\in B^o(x,r)$.
For the second part of Hint 1, set up and simplify $y_n-y$. Here it is true that $y_n-y=(y-x)\frac{1}{n}$, and so $\|y_n-y\|=\|(y-x)\frac{1}{n}\|\to0$, which is exactly what we want, as it shows that $y_n\to y$, i.e. that $y\in\overline{B^o(x,r)}$, as desired.