I wonder about the following statement:

Unit ball is compact in a real Norm Linear Space (NLS) iff the space is finite dimensional.

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

  • 5
    $\begingroup$ NLS=? Moreover, are you sure you don't want to speak of compactness of the unit ball? $\endgroup$ – Giuseppe Negro Jan 26 '13 at 15:22
  • 1
    $\begingroup$ Does "finite" mean "finite-dimensional"? Note that "closed + bounded" does not imply "compact" in infinite-dimensional normed spaces. $\endgroup$ – Martin Jan 26 '13 at 15:27
  • $\begingroup$ Related $\endgroup$ – Pedro Feb 3 '17 at 16:10

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.

  • 1
    $\begingroup$ 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. $\endgroup$ – Martin Jan 26 '13 at 15:38
  • $\begingroup$ @Martin Touche - thank you. $\endgroup$ – Neal Jan 26 '13 at 15:40

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.


It depends what you mean by "unit ball". To be make the question precise, let $$U:=\{x\in X: \|x\| < 1\}$$ and $$E:=\{x\in X: \|x\| \leq 1\}.$$ Then the conclusions are the following;

(1) $U$ is open in $X$, and not closed in $X$.

(2) $E$ is closed in $X$ and not open in $X$.

(3) $E$ is compact if and only if $X$ is finite dimensional.

(4) $U$ is totally bounded in $X$ if and only if $X$ is finite dimensional.

(5) $E$ is complete in $X$ if and only if $X$ is complete.

The proofs of (1), (2) are easy, and is left to the reader as exercises.

For the proofs of (3) and (4) see the book: "Functional Analysis: A First Course", PHI-learning, New Delhi, 2002 (Fourth Print: 2014), by me (M.Thamban Nair).

Proof of (5) is also easy, recently observed by me, and is going to appear in an educational news letter. This shows that if the space is not complete, then its "closed unit ball" $E$ is not complete, and hence not compact.


Is this statement true?

If you are talking about the closed unit ball, yes (as explained in the other answers).

How would I go about proving this?

The proof (of the hardest part) depends on your definition of compactness. If you adopt the sequential definition, the proof follows as explained in the other answers. If you adopt the covering definition, the proof follows as explained below. In the following, $X$ is a normed linear space and $B_X$ stands for the closed unit ball in $X$.

Claim. $B_X$ is compact iff $\dim X<\infty$.


  • ($\Rightarrow$) Assume that $B_X$ is compact.

As $\{B(x;\frac{1}{2})\}_{x\in B_X}$ is an open cover of $B_X$, there are $x_1,...,x_n\in B_X$ such that $$B_X\subset B(x_1;\tfrac{1}{2})\cup \cdots \cup B(x_n;\tfrac{1}{2}).$$ As $B(x_i;\tfrac{1}{2})= x_i+\tfrac{1}{2}B(0;1)$ it follows that $$B(0;1)\subset B_X\subset Y+\tfrac{1}{2}B(0;1),$$ where $Y=\text{span}\{x_1,...,x_n\}$. Thus \begin{align*} B(0;1)\subset Y+\tfrac{1}{2}[Y+\tfrac{1}{2}B(0;1)]&=Y+\tfrac{1}{2^2} B(0;1)\\ &\subset Y+\tfrac{1}{2^2} [Y+\tfrac{1}{2}B(0;1)]=Y+\tfrac{1}{2^3}B(0;1). \end{align*} In general, $$B(0;1)\subset Y+\tfrac{1}{2^n} B(0;1)=Y+B(0,\tfrac{1}{2^n}),\quad \forall\ n\in\mathbb N.$$ So, each $x\in B(0;1)$ has the form $x=y_n+x_n$ with $y_n\in Y$ and $x_n\in B(0;\tfrac{1}{2^n})$. As $x_n\to 0$, we conclude that $y_n\to x$ and thus $x\in \overline{Y}=Y$. This shows that $B(0;1)\subset Y$ which implies that $X\subset Y$ and thus $$\dim X\leq\dim Y=n<\infty.$$

  • ($\Leftarrow$) Assume that $\dim X=n<\infty$.

There is a topological isomorphism $f:X\to \mathbb K^n$. As $B_X$ is closed and bounded, $f(B_X)$ is closed and bounded. As $\mathbb K^n$ has the Heine-Borel property, $f(B_X)$ is compact. Thus $B_X=f^{-1}(f(B_X))$ is compact. $\square$


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.


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