Language/notation abuse when considering an open set of the product topology

Question

Recently, I posted this proof for an exercise in Munkres' Topology book. Now, notice that in the first part, when I consider an open set $U$ from the product topology of some product space $\displaystyle{\prod_{\alpha \in I} X_\alpha}$, I mention that it contains a basis element $B$ such that: $$B = \displaystyle{\prod_{\alpha \in I} U_\alpha}$$ where $U_\alpha$ is an open set of $X_\alpha$ and $U_\alpha = X_\alpha$ for all but finitely many values of $\alpha$.

I know there's nothing wrong with this, but today in class my professor was doing a proof and the first step was taking an open set $U$ from some product space and showing that it satifies some property. Now, the way he wrote it was:

$$U = \displaystyle{\prod_{\alpha \in I} U_\alpha}$$

where $U_\alpha$ is an open set of $X_\alpha$ and $U_\alpha = X_\alpha$ for all but finitely many values of $\alpha$. I asked:

Shouldn't we say that $U$ contains a basis element of that form, rather than $U$ is of this form?

The answer was: it's a basic open set anyhow, so it doesn't make a difference.

Now, here's what I imagine that meant: every property that basis elements have will be preserved to all open sets, since they are unions of basis elements, so anytime we want to prove that something holds for an arbitrary open set, we can just prove that it holds for an arbitrary basis element. But still, it is not entirely correct to say that any open set $U$ is of that form, is it? I know it's probably because it's rather boring to have to write "$U$ contains a basis element of this form" every time, but we could just say "any basic open set $U$ is of this form" (and I wouldn't have minded if that had been what he wrote, but it wasn't) instead of writing "any open set $U$ is of this form", which (I think) is not actually correct. I have looked around and other people do this too and aren't careful enough to write "basic open set" instead of just "open set".

I hope my doubts are clear by now. So, am I right or is there something I'm missing?

Answer 1

You need to be more careful, and I think your professor does too.

every property that basis elements have will be preserved to all open sets, since they are unions of open sets, so anytime we want to prove that something holds for an arbitrary open set, we can just prove that it holds for an arbitrary basis element.

This is wrong, in general. There are certainly properties that hold for all basic open sets, but do not hold for all open sets. As a trivial example, let $U_0$ be your favorite non-basic open set, and consider the property "not equal to $U_0$." (I can't think of a better example off the top of my head, but maybe I will later.)

My guess is that for the statement your professor was trying to prove, say "all open sets have property P", it happens, for this particular statement P, that once you have shown all basic open sets have property P, it is then easy to show that all open sets have property P. (In other words, P is perhaps some property that is "obviously" closed under taking unions.) So one might say something like "it suffices to assume U is a basic open set" or "without loss of generality, suppose U is of the form...", without explaining the step where you extend it to all open sets (because it's simple). I suspect your professor meant this, but explained it poorly.

But this will not necessarily be the case for every possible property P.

Answer 2

In many cases it is actually enough to consider basic open sets instead of general ones, but there is always some small proof that justifies this:

Let $\mathcal{B}$ be a base for some topology $(X, \mathcal{T})$.

• A function $f: X \to Y$ is open iff for every $B \in \mathcal{B}$, $f[B]$ is open in $Y$. (this is true as unions are preserved by forward images).

• A function $f : Y \to X$ is continuous iff $f^{-1}[B]$ is open in $Y$ for all $B \in \mathcal{B}$. (inverse images also preserve unions).

• subset $D$ of $X$ is dense in $X$ iff every non-empty $B \in \mathcal{B}$ intersects $D$. (because every non-empty open set contains a non-empty basic open set).

• The space $X$ is compact iff every open cover by members of $\mathcal{B}$ has a finite subcover. (this is quite simple to see).

• the topology $\mathcal{T}'$ on $X$ is finer than $\mathcal{T}$ iff every $B \in \mathcal{B}$ is in $\mathcal{T}'$, etc.

and many more. The special form of the base elements can then make some proofs a little easier. But you're right that one cannot always do this, you have to stay critical and see if the reduction to considering basic sets only is justified.