Topological properties of a normed space X are widely different depending on X is either finite-dimensional or infinite-dimensional; in fact, whole books are devoted to topology in infinite dimensions. Here, we propose some surprising but elementary facts in infinite-dimensional topology.

Lemma 1: Let X be a normed space and A \subset X be a proper closed linear subspace. For every \epsilon>0, there exists x \in X such that \|x\|=1 and d(x,A) \geq 1- \epsilon.

Proof. Let \epsilon>0; without loss of generality, we may suppose \epsilon<1. Let y \in X \backslash A and a_0 \in A be such that

\displaystyle d(y,A) \leq \| y-a_0 \| \leq \frac{d(y,A)}{1- \epsilon}.

Let \displaystyle x = \frac{y-a_0}{\|y-a_0\|}. Then for all a \in A,

\displaystyle \|x-a\|= \frac{y- (a_0+ \|y-a_0\| \cdot a)}{\|y-a_0\|} \geq \frac{d(y,A)}{\|y-a_0\|} \geq 1- \epsilon,

because a_0+ \|y-a_0\| \cdot a \in A, hence d(x,A) \geq 1- \epsilon. \square

Theorem 1: Let X be a normed space. The unit closed ball B= \{x \in X \mid \|x\| \leq 1\} is compact if and only if X is finite-dimensional.

Proof. Let x_0 \in S(0,1). Because any finite-dimensional linear subspace of X is closed, we can use lemma 1 to construct by induction a sequence (x_n) satisfying

\|x_n\|=1 and d(x_n, \mathrm{span}(x_0,\dots,x_{n-1})) \geq \frac{1}{2}

for all n \geq 1; the sequence has infinitely many terms because X is infinite-dimensional.

In particular \|x_n-x_m\| \geq \frac{1}{2} when n \neq m. Therefore, (x_n) is a sequence in the sphere S(0,1) \subset B without converging subsequence: B cannot be compact. \square

Corollary: Let X be an infinite-dimensional normed space. Then the interior of any compact subspace K \subset X is empty.

Theorem 2: Let X be an infinite-dimensional normed space and K \subset X be a compact subspace. Then X \backslash K is path connected.

Proof. Without loss of generality we may suppose that K \subset B(0,1) since x \mapsto a \cdot x is a homeomorphism for all a \in \mathbb{R} \backslash \{0\}. Because the sphere S(0,1) is path-connected, it is sufficient to show that any point x_0 \notin K can be connected to S(0,1). If \|x_0\|>1 the statement is clear, so from now on suppose that x \in B(0,1).

As in the proof of theorem 1, there exists a sequence (x_n) in S(0,1) satisfying \|x_n-x_m\| \geq \frac{1}{2} for all n \neq m.

Suppose by contradiction that every line \mathbb{R} \cdot x_n -x meets K, that is for every n \geq 1 there exists \lambda_n \in \mathbb{R} such that \lambda_n \cdot x_n-x \in K.

Because K is compact and x \notin K, there exists \delta>0 (independent on n) such that \delta \leq |\lambda_n| \leq 1. Therefore, there exists a subsequence (\lambda_{\sigma(n)}) converging to some \lambda \neq 0.

Again by compactness, the sequence (\lambda_{\sigma(n)} \cdot x_n-x) has a subsequence (\lambda_{\sigma \circ \mu(n)} \cdot x_{\mu(n)}-x) converging to some y \in K.

Therefore, the sequence (x_{\mu(n)}) converges to \frac{1}{\lambda} \cdot (y-x), a contradiction with \|x_n-x_m\| \geq \frac{1}{2} when n \neq m. \square

Theorem 3: For all n \geq 2, the n-dimensional sphere \mathbb{S}^n is simply connected but not contractible.

According to the following lemma, our theorem is just a consequence of Brouwer’s fixed point theorem, proved in a previous post about topological degree theory:

Lemma 2: Let X be a normed space. The following statements are equivalent:

  1. The unit sphere \mathbb{S} is not contractible,
  2. Any continuous map \mathbb{B} \to \mathbb{B} has a fixed point,
  3. \mathbb{S} is not a retract of the unit ball \mathbb{B}.

Proof. Suppose that there exists a continuous map f : \mathbb{B} \to \mathbb{B} without fixed point. Then

H : (t,x) \mapsto \left\{ \begin{array}{cl} \displaystyle \frac{x-2tf(x)}{\|x-2tf(x)\|} & \displaystyle \text{if} \ 0 \leq t \leq \frac{1}{2} \\ \displaystyle \frac{(2-2t)x-f((2-2t)x)}{\|(2-2t)x-f((2-2t)x)\|} & \displaystyle \text{if} \ \frac{1}{2} \leq t \leq 1 \end{array} \right.

defines a homotopy between \mathrm{Id} : \mathbb{S} \to \mathbb{S} and x \mapsto \frac{f(0)}{\|f(0)\|}, so \mathbb{S} is contractible. We proved (1) \Rightarrow (2)

Suppose that there exists a retraction r : \mathbb{B} \to \mathbb{S}. Then f : x \mapsto -r(x) has no fixed point. Indeed, if x_0 \in \mathbb{B} were a fixed point, we would have x \in - r(\mathbb{B})=\mathbb{S} hence r(x)=x whereas r(x)=-x by assumption; therefore, x=0 a contradiction with x \in \mathbb{S}. We proved (2) \Rightarrow (3).

Suppose that there exists a homotopy H : [0,1] \times \mathbb{S} \to \mathbb{S} between \mathrm{Id} : \mathbb{S} \to \mathbb{S} and a constant map H(0, \cdot) : x \mapsto x_0. Then

r : x \mapsto \left\{ \begin{array}{cl} x_0 & \displaystyle \text{if} \ \|x\| \leq \frac{1}{2} \\ \displaystyle H \left( 2 \|x\| -1, \frac{x}{\|x\|} \right) & \displaystyle \text{if} \ \|x\| \geq \frac{1}{2} \end{array} \right.

defines a retraction of \mathbb{B} onto \mathbb{S}. We proved (3) \Rightarrow (1). \square

In order to opposite our theorem with a statement in infinite dimension, we want to introduce an infinite-dimensional sphere \mathbb{S}^{\infty}. A possible construction is the following:

Defintion: Let \mathbb{R}^{\infty} be the direct sum of countably many copies of \mathbb{R} endowed with the norm \displaystyle \| \cdot \| : (x_1,x_2, \dots) \mapsto \left( \sum\limits_{i \geq 1} x_i^2 \right)^{1/2}. Then we define \mathbb{S}^{\infty} = \{x \in \mathbb{R}^{\infty} \mid \|x\|=1\}.

Theorem 4: The infinite-dimensional sphere \mathbb{S}^{\infty} is contractible.

Proof. First, with

H_1 : (t,x) \mapsto (1-t)(x_1,x_2, \dots)+t(0,x_1,x_2, \dots),

the map \displaystyle \frac{H_1}{\|H_1\|} defines a homotopy between the identity \mathrm{Id} : \mathbb{S}^{\infty} \to \mathbb{S}^{\infty} and the shift operator S : (x_1,x_2,\dots) \mapsto (0,x_1,x_2,\dots). Then, with

H_2 : (t,x) \mapsto (1-t) (0,x_1,x_2,\dots)+t(1,0,0, \dots),

the map \displaystyle \frac{H_2}{\|H_2\|} defines a homotopy between S and the constant map x \mapsto (1,0,\dots).

Therefore, \mathrm{Id} : \mathbb{S}^{\infty} \to \mathbb{S}^{\infty} is homotopic to x \mapsto (1,0,\dots), ie. \mathbb{S}^{\infty} is contractible. \square

To understand theorem 4, we may notice that \mathbb{S}^{\infty}= \bigcup\limits_{n \geq 1} \mathbb{S}^n where \mathbb{S}^n is included into \mathbb{S}^{n+1} thanks to (x_1, \dots, x_n) \mapsto (x_1, \dots, x_n,0); in particular, \mathbb{S}^n is a subcomplex of \mathbb{S}^{n+1}. Then, because attaching a m-cell to a CW complex does not change the n-th homotopy group when m>n, we deduce that

\pi_n(\mathbb{S}^{\infty})= \pi_n(\mathbb{S}^{n+1})=0,

that is \mathbb{S}^{\infty} is weakly contractible. However, according to Whitehead theorem, a weakly contractible CW complex is in fact contractible, so \mathbb{S}^{\infty} is contractible.

Therefore, intuitively, we add cells to \mathbb{S}^1 to kill successively the homotopy groups in order to make \mathbb{S}^{\infty} weakly contractible and finally contractible.