Why are facts proven in the (infinite) product topology also true for finite products?

Question

Context question: ZFC

Let $(X_i, \mathcal{T}_i)_{i \in I}$ be a family non empty-topological spaces.

We define the product space as the topological space $(\prod_{i \in I} X_i, \prod_{i \in I} \mathcal{T}_i)$ in the following way:

$\prod_{i \in I} \mathcal{T}_i$ is an initial topology for the source $(\operatorname{pr}_k: \prod X_i \to X_k: f := (f(i))_{i \in I}\mapsto f(k))_{k \in I}$, i.e. it is the smallest topology on the product that makes this source continuous.

Now, this definition works for finite products too. And if $(X, \mathcal{T}),(Y, \mathcal{S})$ are two topological spaces, we define the product topology of the two spaces as the set which has $\mathcal{T} \times \mathcal{S}$ as basis.

Now, clearly there is a difference between the finite product of such two topological spaces, and the product topology as defined above: the latter consists of functions.

So why is it sufficient to prove results in the product topology (as defined first in this post) to prove theorems in the other case? (and vice versa)

Answer 1

Both kinds of products, finite or infinite, consist of functions, really. We see points in $X \times Y$ as functions $f: \{0,1\} \to X \cup Y$ such that $f(0) \in X$ and $f(1) \in Y$, (often denoted $(f(0), f(1))$) but we also see it as a pair $(x,y)$ with $x \in X, y \in Y$. These are equivalent ways to see the finite product. (there is a natural bijection between both representations). The function representation is actually easier (e.g. if we use "pairs" for products of two spaces, we need to prove another bijection between $(X \times Y) \times Z$ and $X \times (Y \times Z)$ to justify the notation $X \times Y \times Z$ as well) and more homogeneous. A finite product is just a product (in the function view) where $I$ happens to be finite.

As you say, everything we can prove for arbitary products using the fact that we have an initial topology induced by the projection maps, a fortiori holds for finite products as well.

"Vice versa" does not always hold: a finite product of locally compact spaces is locally compact but this fails for infinite products. A countable product of first (second) countable spaces is first (second) countable, but this no longer holds for uncountable products, etc.

But indeed the most often used facts about the product topology:

• A product of compact spaces is compact,
• A product of (path-)connected spaces is (path-)connected,
• A product of $T_0$ ($T_1$, Hausdorff, regular) spaces is $T_0$ ($T_1$, Hausdorff, regular),

do hold regardless of the number of factor spaces. A more important distinction often is countable vs uncountable products, as I showed above.

Answer 2

The definition of a product of spaces indexed by I is an infinite product if I is infinite and a finite product if
I is finite. In other words, it is the same definition.

So what is proved for products, is prove for infinite products
and finite products. Finite products however have some
properties infinite products do not have.

Bases of finite products are simply products of any open sets.
Bases of infinite products more complex.