# Is it true that the unit ball is closed in a normed linear space iff the space is finite-dimensional?

I wonder about the following statement:

Unit ball closed in a NLS iff the space is finite.

Is this statement true? How would I go about proving this? I don't want to use that every norm in a finite dimensional NLS is equivalent because I am using the previous result to prove the later.

-
NLS=? Moreover, are you sure you don't want to speak of compactness of the unit ball? – Giuseppe Negro Jan 26 '13 at 15:22
Does "finite" mean "finite-dimensional"? Note that "closed + bounded" does not imply "compact" in infinite-dimensional normed spaces. – Martin Jan 26 '13 at 15:27

## 3 Answers

The unit ball is always closed in a normed (real or complex) linear space. You may be thinking of compactness:

The unit ball is compact iff the space is finite dimensional.

For the first direction, prove the contrapositive by fixing $\epsilon > 0$ and then choosing a countable set of linearly independent vectors with length less than $1$, so that each subsequent vector is further than $\epsilon$ from the span of the previous vectors. This can be done with Riesz's lemma. This is a sequence with no convergent subsequence. For the converse direction, pick a basis and consider $\mathbb{R}^n$.

Thank you for the correction, Martin.

-
Linearly independent is not good enough. Consider $\sqrt{1-2^{-2n}} e_1 + 2^{-n} e_n$ in $\ell^2$. You need to ensure that they are far away from each other, which is usually done by using Riesz's lemma. – Martin Jan 26 '13 at 15:38
@Martin Touche - thank you. – Neal Jan 26 '13 at 15:40

The unit ball is always closed. What is true is that for a normed linear space, the unit ball is compact iff the space is finite dimensional.

-

I assume you are talking about compactness.

($\Leftarrow$) Use criterion of compactness for finite-dimensional subspaces - id est closedness and boundedness.

1.1. Unit ball $B$ of normed space $(X,\Vert\cdot\Vert)$ is bounded by definition of boundedness.

1.2. Unit ball is closed because it is preimage of the closedd set $[0,1]\subset \mathbb{R}$ under the continuous map $$f:X\to\mathbb{R}_+:x\mapsto\Vert x\Vert$$

($\Rightarrow$) Prove ad absurdum. Use Riesz lemma, to construct a sequence $\{x_n:n\in\mathbb{N}\}\subset B$ with the property $$\forall n,m\in\mathbb{N}\qquad n\neq m\implies \Vert x_n-x_m\Vert>1/2$$ Show that it have no convergent subsequence.

-