What is wrong with the example I gave to contradict a theorem that claim that the closure of $B$ is the unit ball?

Question

I am studying functional analysis, and I saw the following claim

Let $X$ be a vector space, $B\subseteq X$ is convex, symmetric around $0$ and s.t $$\bigcup_{n=1}^{\infty}nB=X,\quad\bigcap_{n=1}^\infty \frac{1}{n}B=\{0\}$$ then $$\|x\|:=\inf\{\lambda>0\mid\frac{x}{\lambda}\in B\}$$ is a norm on $X$, moreover, the unit ball satisfy $\overline{B}=B_{1}$

Why, for example, isn't $$X=\mathbb{R}^{2},\, B:=\{x\in\mathbb{R}^{2}\mid||x||\leq2\}$$a counter example to the claim that $\overline{B}=B_{1}$ ?

It looks as if it satisfies all assumptions, but not the conclusion.

The problem means that the unit ball in the norm $||x||$ satisfies $\overline{B}=B_1$. — KReiser, Oct 01 '13 at 17:47
The norm you use to define $B$ is not the norm $\lVert\cdot\rVert_B$ that $B$ induces. — Daniel Fischer, Oct 01 '13 at 17:47

score 2 · Accepted Answer · answered Oct 01 '13 at 17:48

2

Because the new norm you get is NOT $\| \|_2$.

Note that in this example:

$$\| x \|:= \inf\{\lambda>0\mid \| \frac{x}{\lambda}\| \leq 2 \} =2 \| x\|_2$$

And with respect to this norm, $B_2$ becomes the unit ball...

answered Oct 01 '13 at 17:48

N. S.

132,525

What is wrong with the example I gave to contradict a theorem that claim that the closure of $B$ is the unit ball?

1 Answers1