Proving $(\mathbb{R},\mathcal{T}_{cf}), (\mathbb{R},\mathcal{T}_u)$ are not homeomorphic

Question

I´d like to know if it is valid this argument to prove $\mathbb{R}$ with the finite complement topology is homeomorphic to itself with the usual topology:

Argument #1: We take $(a,+\infty) \in \mathcal{T}_u$. Then $f^{-1}((a,+\infty)) \notin \mathcal{T}_{cf}$, because $f^{-1}((a,+\infty))^c = f^{-1}((a,+\infty)^c) = f^{-1}((-\infty,a])$ and, since $f$ is bijective, necessarily $f^{-1}((-\infty,a])$ is not finite. So $f$ continious can´t exist.

Argument #2: In $(\mathbb{R},\mathcal{T}_{cf})$ the only closed subsets are point (or finite union of points) and in $(\mathbb{R},\mathcal{T}_u)$ there are more possibilities. So can´t exist $f$ hoemomorphism such that send closed sets to closed sets.

Argument #3: $\mathcal{T}_{cf} \subset \mathcal{T}_u$, so can´t exist a homeomorphism.

Thanks!!

Answer 1

I think that the second argument is the clearest. If there were a homeomorphism from $(\mathbb{R}, \mathcal{T}_u)\to (\mathbb{R},\mathcal{T}_{cf})$, then it would induce a bijection from each closed set in $\mathbb{R}$ with the standard topology to a closed set in $\mathbb{R}$ with the finite complement topology. This is a contradiction, because the standard topology has infinite closed sets (e.g. $[a,b]$) while the finite complement topology has only finite cardinality closed sets.