# Approximation of homotopy avoiding a point in $\mathbb{R}^3$

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.

• Hint: consider the set of homotopies from the curve to a point that miss a given fixed rational point. – alesia Feb 6 at 22:02
• 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 – Swann Feb 6 at 22:42

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.