Understanding Theorem 19.2 in Munkres

Question

Theorem 19.2 in Munkres Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathcal{B}_{\alpha}$. The collection of all sets of the form $$\prod_{\alpha \in J} B_{\alpha}$$ where $B_{\alpha} \in \mathcal{B}_{\alpha}$ for each ${\alpha}$, will serve as a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$

The problem I'm having with understanding the theorem is due to the use of the indices.

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From what I understand the above theorem is saying that the collection of all possible sets formed by:

Taking one arbitrary basis element from each arbitrary basis $\mathcal{B}_{\alpha}$, and then taking the product of each of those basis elements

forms a basis for the box topology $\prod_{\alpha \in J} X_{\alpha}$.

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Is my understanding correct? I apologize in advance if this question is somewhat vague.

Answer 1

Yes, your interpretation is correct. The open sets in the box topology on $\displaystyle \prod_{\alpha \in J} X_\alpha$ will be unions of basis elements. The basis elements are of the form $\displaystyle \prod_{\alpha \in J} B_\alpha$, where each $B_\alpha$ is some basis element for $X_\alpha$.

Perhaps more simply, the basis for the box topology can be defined in terms of open sets. That is, $\displaystyle \prod_{\alpha \in J} X_\alpha$ has as a basis $\displaystyle \left\{ \prod_{\alpha \in J} U_\alpha \ \Big| \ U_\alpha \text{ is open in } X_\alpha \right\}$.

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As a side note, we typically choose the product topology over the box topology in practice; I'm sure this'll come up shortly in Munkres. The product topology is defined as we defined the box topology in the second paragraph above, but with the added stipulation that $U_\alpha = X_\alpha$ for all but finitely many $\alpha \in J$. Though this seemingly more complicated, the product topology satisfies more "desirable" things. Some examples off the top of my head include:

• The product of compact spaces will be compact.

• A function $f: Y \rightarrow \displaystyle \prod_{\alpha \in J} X_\alpha$ is continuous $\iff$ each $(\pi_\alpha \circ f): Y \rightarrow X_\alpha$ is continuous, where $\pi_\alpha$ denotes the projection map from the product space onto $X_\alpha$.

Lastly, note that the product topology and the box topology are the same thing unless we are looking at the product of infinitely many $X_\alpha$.