# Let $R$ be the set $\mathbb{R}$ of all real numbers with the co-finite topology. How can one show that $R$ is contractible?

I have recently started learning Algebraic Topology, and more specifically, homotopy theory. That is where I encountered this question:

Let $$R$$ be the set $$\mathbb{R}$$ of real numbers, with the co-finite topology. Show that $$R$$ is contractible.

I already have some knowledge of point-set topology, the definition of homotopy, homotopy equivalence and the meaning of a contractible space.

The idea I used to approach the problem was to try and show that the identity map on $$R$$ was null homotopic. I considered the identity map $$id:R\to R$$ and $$c:R\to R$$, defined as $$c(x)=x_0$$ for all $$x\in R$$, where $$x_0\in R$$. I tried to check if the function $$H:R\times\mathbb{I}\to R$$ (where $$\mathbb{I}=[0,1]$$), given by $$H(x,t)=(1-t)x+tx_0$$, was continuous in the product topology by checking if the pre-images of closed sets are closed, but that is where I am stuck. I am finding it hard to do that.

Besides that, I found a result on the following webpage that might be helpful, albeit requiring a different approach. If it is possible to show that $$R$$ is path-connected, then that result suggests that it would be contractible. I am yet to perfectly understand the proof to it, but it could be applied too. Here's the link to it: A strange contractible space.

Any help would be much appreciated.

• – Moishe Kohan Jun 8 at 5:01
• @MoisheKohan I am actually not aware of the Zariski topology. Is it exactly the same as the co-finite topology? Also, how, in that answer, the OP said that there would exist a bijection $\phi: \mathbb{C}^{1} \times (0, 1) \simeq \mathbb{C}^{1}$ and thus the pre-images of points would be closed, isn't really clear to me. – Vishal Agarwal Jun 8 at 5:18
• Yes, in your case Zariski topology is the same as the cofinite topology. If you have a bijection, preimages of points are points, so they are closed. – Moishe Kohan Jun 8 at 5:28
• @MoisheKohan Okay, I see what you are getting at. I now need to convince myself that there does, in fact, exist such a bijection between that product space and the $\mathbb{C}^1$ space. – Vishal Agarwal Jun 8 at 6:43
• $X$ and $X^2$ have the same cardinality if $X$ is infinite. From this, you see that ${\mathbb R}$ and ${\mathbb R}\times (0,1)$ have the same cardinality. – Moishe Kohan Jun 8 at 12:28

Let $$X$$ have the cofinite topology; then if $$X$$ is path-connected, then $$X$$ is contractible.
And $$\Bbb R$$ in the cofinite topology is path-connected: if $$x \neq y$$ in $$\Bbb R$$ then $$p: [0,1] \to \Bbb R, p(t)=tx+(1-t)y$$ is injective and thereby automatically continuous when the codomain has the cofinite topology (inverse images of closed sets are closed!), as $$[0,1]$$ is metric hence $$T_1$$.
So the proof in the blog applies and $$\Bbb R$$ in the cofinite topology is contractible.