This is exercise 2.6(c) from Kosniowski's A first course in algebraic topology:
Show that if a topological space has only a finite number of points each of which is closed then it has the discrete topology
My attempt:
Let $X$ be a topological space as mentioned. If $U$ is a subset of $X$ then it is closed. This implies it's complement is open. The complement is also a subset of $X$ and hence it is also closed. Therefore, double complement of $U$ is open which means $U$ is open.
My question is that why is there a restriction of finite number of points?
Best Answer
How do you know this? This is not true in general. For example, in $\mathbb R$ with the standard topology, there exist many sets which are subsets of $\mathbb R$ but not closed.
In your case, it is indeed true that all subsets of $X$ are closed. But you have to prove that they are closed, you cannot just write that down and hope for the best.
Hint:
To prove that $U$ is closed, you will have to use some specific properties of $X$.
Also, just a general piece of advice. Are you sure you want to tackle algebraic topology if you so far do not yet have a firm grasp on topology? I don't think that's a good idea.