There exist a large number of, sometimes surprising, applications of Baire category theorem. For instance, in this blog, we saw that

• two generic orientation-preserving homeomorphisms of the circle $\mathbb{S}^1$ generate a non abelian free subgroup,
• a generic continuous function $[0,1] \to \mathbb{R}$ is nowhere differentiable.

See the previous notes Free groups acting on the circle and Almost all continuous functions are nowhere differentiable. Now, we want to prove:

Theorem: If $n \geq 3$, the space $\mathbb{R}^n \backslash \mathbb{Q}^n$ is simply connected.

Fix a base point $x \in \mathbb{R}^n \backslash \mathbb{Q}^n$ and an enumeration $\mathbb{Q}^n = \{ q_1, q_2, \ldots \}$. Given two closed curves $c_0,c_1 : [0,1] \to \mathbb{R}^n \backslash \mathbb{Q}^n$ satisfying $c_0(0)=c_1(0)=x$, we want to prove that $c_0$ and $c_1$ are homotopic (relatively to $x$) in $\mathbb{R}^n \backslash \mathbb{Q}^n$. Let

$\mathcal{H}(c_0,c_1) = \{ \text{homotopy between} \ c_0 \ \text{and} \ c_1 \ \text{in} \ \mathbb{R}^n \}$.

It is a closed subset of the Banach space $\left( \mathcal{C}^0([0,1]^2, \mathbb{R}^n), \| \cdot \|_{\infty} \right)$ of the continuous functions $[0,1]^2 \to \mathbb{R}^n$ with respect to the supremum norm $\| \cdot \|_{\infty}$, so that $\mathcal{H}(c_0,c_1)$, endowed with the distance induced by $\| \cdot \|_{\infty}$, is a complete metric space. Now, let

$\mathcal{H}^k(c_0,c_1) = \{ h \in \mathcal{H}(c_0,c_1) \mid \mathrm{Im}(h) \cap \{ q_1, \ldots, q_k \}= \emptyset \}$.

Because $n \geq 3$, the space $\mathbb{R}^n \backslash \{ q_1, \ldots, q_k \}$ is simply connected, so that $\mathcal{H}^k(c_0,c_1) \neq \emptyset$. In fact, it is not difficult to convince ourself that $\mathcal{H}^k(c_0,c_1)$ is a dense open subset of $\mathcal{H}(c_0,c_1)$. A possible argument is the following:

First, if $h \in \mathcal{H}^k(c_0,c_1)$ and if $d$ denotes the distance between $\{q_1, \ldots, q_k \}$ and $\mathrm{Im}(h)$ (which is positive since $\mathrm{Im}(h)$ is compact), then the ball in $\mathcal{H}(c_0,c_1)$ of radius $d/2$ and centered at $h$ is clearly included into $\mathcal{H}^k(c_0,c_1)$. Therefore, $\mathcal{H}^k(c_0,c_1)$ is open in $\mathcal{H}(c_0,c_1)$.

Then, let $f_{\epsilon}$ be a bump function whose support is included into $\bigcup\limits_{j=1}^k B(q_i,\epsilon)$ and satisfying $\| f_{\epsilon} \|_{\infty} \leq \epsilon$. Now, given a homotopy $h \in \mathcal{H}(c_0,c_1)$, define the perturbation

$h_{\epsilon} : (s,t) \mapsto (t(t-1)f_{\epsilon} + \mathrm{Id}_X) \circ h(s,t)$.

Of course, $\| h-h_{\epsilon} \|_{\infty} \leq \| f_{\epsilon} \|_{\infty} \leq \epsilon$ and $h_{\epsilon} \in \mathcal{H}(c_0,c_1)$. Furthermore, if $\epsilon$ is small enough and $f_{\epsilon}$ well chosen, then $\mathrm{Im}(h_{\epsilon}) \cap \{ q_1, \ldots, q_k \} = \emptyset$. This proves that $\mathcal{H}^k(c_0,c_1)$ is dense in $\mathcal{H}(c_0,c_1)$.

Now, it is sufficient to apply Baire category theorem to conclude that the intersection $\bigcap\limits_{k \geq 1} \mathcal{H}^k (c_0,c_1)$ is non empty. But any element $h$ of this intersection defines a homotopy between $c_0$ and $c_1$ satisfying $\mathrm{Im}(h) \cap \mathbb{Q}^n = \emptyset$, that is to say a homotopy in $\mathbb{R}^n \backslash \mathbb{Q}^n$.

Therefore, we have proved that any two closed curves in $\mathbb{R}^n \backslash \mathbb{Q}^n$ based at $x$ are homotopic in $\mathbb{R}^n \backslash \mathbb{Q}^n$. This shows that the group $\pi_1(\mathbb{R}^n \backslash \mathbb{Q}^n,x)$ is trivial, ie., the space $\mathbb{R}^n \backslash \mathbb{Q}^n$ is simply connected.