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.
Best Answer
In your linked blogpost there is the lemma
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.