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.