Why does the product topology allow proper subsets for only finitely many elements? Consider Theorem 19.1 from Munkres' topology:

The box topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$. The product topology on $\prod X_\alpha$ has as basis all sets of the form $\prod U_\alpha$, where $U_\alpha$ is open in $X_\alpha$ for each $\alpha$ and $U_\alpha$ equals $X_\alpha$ except for finitely many values of $\alpha$.

I am having a hard time seeing how the last sentence follows from the subbasis-definition of the product topology. I don't find Munkres' preceding explanation to be helpful.
 A: The product topology can be described in a few equivalent ways. One of these ways is "the weakest topology on the set $X=\prod X_\alpha$, that makes all the projections $X\to X_\alpha$ continuous.
Since all the projections need to be continuous, it follows immediately that any subset of the form $\prod U_\alpha$, where $U_\alpha=X_\alpha$ for all $\alpha$ except for one $\alpha$ for which $U_\alpha\subset X_\alpha$ is any open subset, needs to be open. Taking finite intersections of all the subsets of this form yields the basis mentioned in the question, and the topology it generates has the desired property of the product.
Another way to describe the product topology is by the universal property, which is: For any topological space $A$ and continuous maps $f_\alpha:A\to X_\alpha$, there is a unique continuous map $F:A\to \prod X_\alpha$, such that for every $\alpha$ we have $f_\alpha=\pi_\alpha\circ F$.
Now take for example $\mathbb{R}^\mathbb{N}$, the product of countably many copies of the real line. Let $A=\mathbb{R}$, and for every $n\in\mathbb{N}$ let $f_n:A\to\mathbb{R}$ be given by $t\mapsto n\cdot t$. Obviously, there is exactly one way to define $F:A\to\mathbb{R}^\mathbb{N}$ that will satisfy the above property. The topology on the product needs to be such that this map will be continuous. Consider now $U=\prod_{n\in\mathbb{N}}U_n$, where for every $n$, $U_n=(-1,1).$ Obviously, $$F^{-1}(U)=\bigcap_{n\in\mathbb{N}}\left(-\frac{1}{n},\frac{1}{n}\right)=\{0\},$$which is not open. This shows that the above $U$ shouldn't be open in the product topology.
A: There are two ways to view the generation of a topology from a subbasis $\mathcal S$.


*

*The topology generated by $\mathcal S$ is the smallest (coarsest) topology in which all the sets in $\mathcal S$ are open.

*We first transform $\mathcal S$ into a basis $\mathcal B$ consisting of all (nonempty) finite intersections of sets in $\mathcal S$, and then take the topology generated by this basis.
These two ways can be shown to be equivalent, though the second method is somewhat more illustrative in the current situation, since we outright say that the basis open sets are finite intersections of subbasic open sets. This means that $U$ is open in the generated topology if and only if for each $x \in U$ there are $W_1 , \ldots , W_n$ (for some $n \geq 1$) in $\mathcal S$ such that $x \in W_1 \cap \cdots \cap W_n \subseteq U$.
Since the subbasis for the product topology is the family of all sets $\pi_\beta^{-1} ( U )$ where $\beta$ an index, and $U$ is open in $X_\beta$, then the basic open sets are of the form $$\pi_{\beta_1}^{-1} ( U_{\beta_1} ) \cap \cdots \cap \pi_{\beta_n}^{-1} ( U_{\beta_n} ),$$ where $\beta_1 , \ldots , \beta_n$ are indexes, and $U_{\beta_i}$ is open in $X_{\beta_i}$ for all $i$.  We may then note that if $\beta_i = \beta_j = \beta$, then $\pi_{\beta_i}^{-1} ( U_{\beta_i} ) \cap \pi_{\beta_j}^{-1} ( U_{\beta_j} ) = \pi_{\beta}^{-1} ( U_{\beta_i} \cap U_{\beta_j} )$, and $U_{\beta_i} \cap U_{\beta_j}$ is open in $X_\beta$, and so each index need only appear (at most) once.
