which of the following is $T_1$ space?

Question

From Topology without tears

A Topological space $(X,\tau)$ is said to be $T_1$ space if every singleton set is closed in $(X,\tau)$. Show that precisely two of the following 9 topological spaces are $T_1$.

1.a discrete space

(I believe discrete is $T_1$ because every subset in discrete space is open and closed both.So singleton are also closed.)

2.An indiscrete space with atleast two points

(Since its topology contains only $X$ and $\phi$ I believe singleton will not be neither closed nor open.)

3.An infinite set with cofinite topology

(I have a example $\tau$ consisting of $\mathbb{N}$,$\phi$ and every set $\{n,n+1,...\}$ for any positive integer.Here $\mathbb{N}$ is infinite set and this is cofinite topology since every closed set will be finite here.But here singletons need not be closed.)

4.$X=\{a,b,c,d,e,f\}$ and $\tau=\{ X,\phi,\{a\},\{c,d\},\{a,c,d\},\{b,c,d,e,f\}\}$

(Here I can see every singleton are not closed.)

5.$\tau_1$ consist of $\mathbb{R},\phi$, and every interval $(-n,n)$ for any $n$ positive integer.

(for $\{n\}$ to be closed its complement $(-\infty,-n] \cup [n,\infty)$ must belong to $\tau_1$.Which is not the case.So $\tau_1$ is not $T_1$)

6. $\tau_2$ consist of $\mathbb{R},\phi$, and every interval $[-n,n]$ for any $n$ positive integer.

(Similarly I can show $\tau_2$ not $T_1$)

7.$\tau_3$ consist of $\mathbb{R},\phi$, and every interval $[n,\infty)$ for any $n$ positive integer.

(Similarly $\tau_3$ is also not $T_1$.

8.$\tau_1$ consists of $\mathbb{N}$,$\phi$ and every set $\{1,2,...,n\},$for $n$ any positive integer.

(Here also all singleton will not be closed.because its complement need not be in $\tau_1$)

9.$\tau_2$ consists of $\mathbb{N}$,$\phi$ and every set $\{n,n+1,...\},$for $n$ any positive integer.

(Here also all singleton will not be closed.because its complement need not be in $\tau_1$)

So,I got only one $T_1$ space.where I am wrong?

Thanks in advance!

Answer 1

$3$ is also a $T_1$ space. The complement of singleton $\{a\}$ is $X-\{a\}$ which is open since its complement $\{a\}$ is finite (a non-empty set is open in $\color{red}{\text{co}}$finite topology iff its complement is finite).

The example you have taken is not the cofinite topology on $\Bbb N$. It is a subset (strictly coarser) of the cofinite topology. Note that $\Bbb N-\{2\}\in$ the cofinite topology on $\Bbb N$ but not your example topology.