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$\{X_{\alpha}\}_{\alpha\in \Lambda}$ be discrete topological spaces and $X=\prod_{i\in\Lambda} X_i.$ Then which of the following statements imply that the product topology on $X$ equals the discrete topology on $X\ ?$

$1.\Lambda \text{ is finite}.$

$2.\Lambda \text{ is countably infinite and }X_{i} \text{ are singletons for all but a finite number of } i's.$

$3.\Lambda \text{ is uncountably infinite and }X_{i} \text{ are singletons for all but a finite number of } i.$

$4.\Lambda \text{ is infinite and }X_i \text{ are infinite for all }\ i.$

For case $1$, product and discrete topology on $X$ are same.

For case $2$, let the ones that are not singletons be names $X_1,X_2,............X_n,$ and the rest are $\{x_{n_1}\},\{x_{n+2}\}...............$ and so on. Here too both the topologies are same follows from the definition of product topology.

For case $3$, since only finitely many of the $X_i$'s have more than one element, product and box topology are same on this too.

Now case $4$ is confusing me a little but I think this will have both the topologies same since arbitrary union of open sets is open.

Am I correct ? Please correct me if there is any mistake and help find the right answers. Thanks.

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user118494
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1 Answers1

4

Recall that a basis for $\prod_{i\in\Lambda}X_i$ is given by the sets $\prod_{i\in\Lambda}U_i$ where $U_i\subseteq X_i$ are open and all but finitely many are the whole of $X_i$. It is easy to see that this implies that for each open set $U\subseteq\prod_{i\in\Lambda}X_i$ we have that $\pi_i(U) = X_i$ for infinitely many $i\in\Lambda$, where $\pi_i$ is the projection onto $X_i$ (to see this, notice that this is true for the elements of the basis and remains true when taking arbitrary unions and finite intersections). Therefore, the topology on $\prod_{i\in\Lambda}X_i$ cannot be discrete in case (4).

As remarked by @BrianM.Scott in the comments, this in fact shows that if infinitely many $X_i$ have strictly more than one element, then $\prod_{i\in\Lambda}X_i$ cannot have the dicrete topology.

You found the correct answer for the other three cases.

Daniel Robert-Nicoud
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