Let $x \in \mathbb{R}$. The finite complement topology on $\mathbb{R}$ is a collection of subsets of $U$ of $\mathbb{R}$ such that $U^{c}$ is finite or all of $\mathbb{R}$. Since $\{x\}^{c}$ is infinite in $\mathbb{R}$, $\{x\}$ is not open (not necessarily closed). This question arises as we were given the task of giving an example of a topological space that is not Hausdorff but does satisfy the $T_{1}$ axiom, that finite point sets are closed. The finite complement is not Hausdorff as given any $x_{1},x_{2} \in \mathbb{R}$, the only neighborhood in the finite complement topology which contains either point is all of $\mathbb{R}$. Thus, there does not exists neighborhoods $U_{1},U_{2}$ of $x_{1},x_{2}$ that are disjoint. So finally I come back to my question, is $\{x\}$ closed in the finite complement topology? My proof began claiming that it is not open (but not necessarily closed), however the solution I am looking at for this problem claims that the finite complement topology is not Hausdorff, for which I agree, and also satisfies the $T_{1}$ axiom, hence $\{x\}$ closed, which I have yet to prove. Since I am new to this, I am making an educated guess that there must be another way to prove that $\{x\}$ is closed or I am misinterpreting the meaning of open/closed sets in the finite complement topology. I would appreciate if someone would clear this up for me. Thank you in advance.
[Math] In the finite complement topology on $\mathbb{R}$, is the subset $\{ x \}$ closed
general-topology
Best Answer
The finite complement topology on any set is $T_1$. Let $x$ and $y$ be distinct points. Then $X\setminus\{y\}$ is an open nbhd of $x$ that does not contain $y$, and $X\setminus\{x\}$ is an open nbhd of $y$ that does not contain $x$. Here I’ve used what I consider the most usual definition of the $T_1$ property. If your definition of the $T_1$ property is that singleton sets are closed, you can adapt the argument very easily: for any $x\in X$, $X\setminus\{x\}$ is open, so its complement, $\{x\}$, is closed.