# Is $\tau=\{\mathbb{R},\emptyset$ and every interval $(-n,n)\}$ a $T_1$ -space

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.

Question:

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

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.