Different ways to state the motivation of the definition of the product topology

Question

Suppose for every $i\in\mathscr I,$ $X_i$ is a topological space.

The product space has as its underlying set the product set $X =\prod \limits_{i\,\in\,\mathscr I} X_i$ and as its open sets product sets of the form $\prod\limits_{i\,\in\,\mathscr I} G_i$ where for every $i\in\mathscr I,$ $G_i$ is open and for all except finitely many $i\in\mathscr I,$ $G_i=X_i.$

Now suppose one is asked why the definition is that rather than something else ‒ for example, omitting the restriction to finitely many factors.

The answer that I know instantly is this: This is the same as the topology of pointwise convergence. That is, a net of points in $X$ converges to a point in $X$ if and only if for every $i\in\mathscr I,$ the projection of the net onto the $i$th factor space is a net that converges to the projection of the limit point onto that factor space.

However, there may be other and maybe even better ways of stating the motivation. What are they?

Answer 1

In any category (such as the category Set of sets, the category Grp of groups, ...), the product of objects $A_i$, $i\in I$, is an object $P$, together with morphisms (called canonical projections) $\pi_i\colon P\to A_i$ such that for every object $X$ and family of morphism $f_i\colon X\to A_i$, there exists one and only one morphism $h\colon X\to P$ such that $\pi_i\circ h=f_i$ for all $i\in I$.

Spelling this out for the category Top of topological spaces, leads to the well-known concrete construction.

Answer 2

The product topology is the coarsest topology for which each projection $\pi_i : X \rightarrow X_i$ is continuous.

This makes the product topology the categorical product in the category of topological spaces. That is, for any other space $Y$ with maps $f_i : Y \rightarrow X_i$ we get a unique map $f: Y \rightarrow X$ such that $\pi_i \circ f = f_i$.

Answer 3

The two biggest reasons I can think of for why the product topology is defined this way:

• Continuous functions Given continuous functions $f_i:Y\to X_i$ for each $i\in I$, there is a unique continuous function $f:Y\to\prod_{i\in I}X_i$ such that $f_i=p_i\circ f$ for all $i\in I$, where $p_i:\prod_{j\in I}X_j\to X_i$ is the projection map. In fact, this uniquely characterizes the product topology.
• Compact sets If each $X_i$ is compact, then Tychnoff's theorem states that $\prod_{i\in I}X_i$ is compact. This is an extremely powerful theorem, and is equivalent to the axiom of choice. This is certainly not true under what some may see as a more "natural" topology on $\prod_{i\in I}X_i$, i.e. the topology generated by $\{\prod_{i\in I}U_i:U_i\text{ open for each }i\in I\}$.

Answer 4

For the sake of definiteness, I will refer to the name I have seen most commonly used: the product topology is, as others have mentioned, the Initial Topology https://en.wikipedia.org/wiki/Initial_topology with respect to the projections.

A dual concept is that of the Final Topology, which is the finest topology put on the codomain in $f: Y \rightarrow X $ that makes the set continuous.

https://en.wikipedia.org/wiki/Final_topology.

As an example, the quotient topology is the final topology with respect to quotient maps.

Answer 5

I note that the given definition seems to avoid invoking Choice -- I only have to provide an open set for a finite number of indices. This might be a red herring. (I would be interested if anyone has references that shed light on this thought.)

Edit

Since there seem to be at least two readers who believe that ZF is capable of picking out elements of infinite Cartesian products of nonempty sets, more detail...

Let $\mathscr{I}$ be countably infinite, and let $X_i = X$, a nonempty topological space, for all $i \in \mathscr{I}$. Let $T$ be the topology of $X$. Then $P = $"$\prod_{i \in \mathcal{I}} X_i$" ($ = X^\omega)$ and $Q = $"$\prod_{i \in \mathscr{I}} T $" ($ = T^\omega$) are wffs that ZF cannot prove are sets. ZF can prove arbitrary finite Cartesian products are sets, but has no mechanism to identify an infinite Cartesian product as a set. So, in particular, ZF cannot apply Foundation to $Q$, a basis of the box topology on whatever $P$ is, to get an arbitrary element of $Q$. As soon as one adds a suitable axiom, for instance, the axiom of countable choice, these obstructions vanish.

However, let $Q' = $"$\prod^{(\text{finite})}_{i \in \mathscr{I}} T $", where the superscript "(finite)" means this is the collection of maps from $\mathbb{N}$ to $T$ where, for each member of the collection, only finitely many elements of $\mathbb{N}$ have an image other than $X$. Since finite choice is a theorem of ZF, we are able to construct the set of members of $Q$ with only the first $0$ elements of $\mathbb{N}$ not mapped to $X$, the set of members of $Q$ with only the first $1$ elements of $\mathbb{N}$ not mapped to $X$, and so forth by induction. Via Comprehension, these sets can be assembled into the images of another function from $\mathbb{N}$, allowing ZF to show that $Q'$ is a set. Since $T$ is nonempty, we know that the image of $1$ under this new map is nonempty. Now ZF's Foundation permits "let $U \in T^\omega$".

So the product topology can be framed entirely in the setting of ZF (for countable products of spaces). But the box topology already requires a choice axiom for the smallest infinite Cartesian product.