# Is the product topology the same as the discrete topology on $\prod_{n\in\mathbb{N}}A_{n}$?

Let $\{A_{n}\}_{n\in\mathbb{N}}$ be a countably infinite collection of finite sets, such that each $A_{n}$ has more than one element. Give each $A_{n}$ the discrete topology, and consider the associated product topology on $\prod_{n\in\mathbb{N}}A_{n}$. Is the product topology the same as the discrete topology on $\prod_{n\in\mathbb{N}}A_{n}$?

I know that they are not the same after some examples myself, but I’m not sure how to go about actually prove it in a mathematical way.

• Well, if you have counterexamples, they suffice for an actual proof that these two topologies are different. – lisyarus Nov 23 '17 at 13:59
• With the product topology, the product is compact (Tychonoff), but with the discrete topology, it is not (why?). – PhoemueX Nov 23 '17 at 15:03
• Note that the usual subbase of the product topology will not contain singletons here. – drhab Nov 23 '17 at 16:00
• lisyarus makes the point that a counterexample suffices to show that they are not always the same. PhoemueX makes the stronger claim that in fact you can prove that they are never the same. – Qiaochu Yuan Nov 23 '17 at 18:09
• @QiaochuYuan Indeed. But the theorem of Tychonoff is quite deep. It can also be proved on what I call an elementary way. – drhab Nov 23 '17 at 21:20

The product topology has a subbase in the collection of sets $\prod_{n\in\mathbb N}B_n$ for which some $N\in\mathbb N$ exists with $n>N\implies B_n=A_n$.

Any open set in the product topology is a union of these sets (eventually an empty union).

That means that singletons cannot be open, since none of the sets $\prod_{n\in\mathbb N}B_n$ is a singleton, and a union of non-singletons cannot be a singleton.

Final conclusion: the product topology is not discrete.

When you want to prove some statement is true, you can prove it by a

• Direct proof