Exercise:A topological space $(X,\tau)$ is said to be $T_1$-space if every singleton set $\{x\}$ is closed in $(X,\tau)$. Show that precisely two of the following nine topological spaces are $T_1$-spaces:

v) $\tau$ consists of $\mathbb{R},\emptyset$ and every interval $(-n,n)$ for any positive integers.

I think that $X\setminus(-n,n)=(-\infty,-n]\cup[n,\infty)$ hence the compliment of each set is not a singleton then no singleton is closed.


Is my argument valid? Can I get singletons in the topology(using set operations)?

Thanks in advance!

  • $\begingroup$ @bof It was a mistake of mine. Thanks for pointing that out $\endgroup$ – Pedro Gomes Sep 28 '18 at 11:02

Your argument is correct: you have described all closed sets (other than $\mathbb{R}$ and $\emptyset$) and none of them is a singleton. Therefore, no singleton is closed here.


You can't get the singletons as closed sets. Suppose you can. Take $\{0\}$ and say it is closed. So the complement $U=(-\infty,0)\cup(0,\infty)$ must be open. It means every point in $U$, for example $1$, must be inner point. Than we have to find some $n$ such that $1\in (-n,n)\subset U$. So $0\in (-n,n)\subset U$ and we get contradiction.


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