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For a proof that $\mathbb{R}^3\setminus \mathbb{Q}^3$ is simply connected using Baire category theorem I need to approximate an homotopy $H : [0,1]\times \mathbb{S}^1 \to \mathbb{R}^3$ from a loop $\gamma$ to a point $x$ by another avoiding a point $q$ ($\gamma(z) \neq q $ $\forall z\in\mathbb{S}^1$ and $x\neq q$) with an arbitraty error $\varepsilon$. In symbols : $H' : [0,1]\times \mathbb{S}^1 \to \mathbb{R}^3\setminus\{q\}$ such that $H'(0,z) = \gamma(z)$ , $H'(1,z) = x$ and $\underset{(t,z)\in [0,1]\times\mathbb{S}^1}{\sup}\vert H(t,z) - H'(t,z)\vert < \varepsilon$.

I do not even know if it is true.

Also, if you know another proof of the simple connectess of $\mathbb{R}^3\setminus\mathbb{Q}^3$ (if it is simply connected) it would be nice to be able to see it.

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  • $\begingroup$ Hint: consider the set of homotopies from the curve to a point that miss a given fixed rational point. $\endgroup$ – alesia Feb 6 at 22:02
  • $\begingroup$ It's because I want to show that such a set is dense (as it is open) that I want to approximate a given homotopy $\endgroup$ – Swann Feb 6 at 22:42
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The $\epsilon$-approximation $H':[0,1]\times\mathbb{S}^1\to\mathbb{R}^3\setminus \{ q\}$ of $H$ exists.

Proof. You approximate $H$ by $\tilde{H}$ so that in $\tilde{H}\in C^\infty((0,1)\times\mathbb{S}^1)$ and $\tilde{H}=H$ on $\partial([0,1]\times\mathbb{S}^1)$. You simply use approximation by convolution with the size of the kernel going to zero as you approach the boundary. This way $\tilde{H}$ is $\epsilon/2$ close to $H$, but it may still pass through $q$. Since the image of $\tilde{H}$ is two dimensional you can find $\tilde{q}$ very close to $q$ and not in the image of $\tilde{H}$. Then there is an isotopy $I$ of $\mathbb{R}^3$ that maps $\tilde{q}$ to $q$ and is identity outside a small neighborhood that contains $q$ and $\tilde{q}$ so it is identity near the image of $H(\partial([0,1]\times\mathbb{S}^1)=\tilde{H}(\partial([0,1]\times\mathbb{S}^1)$. The mapping $H'=I\circ\tilde{H}$ is the homotopy you are looking for.

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