0
$\begingroup$

For example, how do we prove that $(-1, 1)^\omega$ is not open in $\mathbb{R}^\omega$? I understand the general form of an open set in this topology, and it's difference with the box topology. The problem is that, while it seems intuitively evident, I'm not sure how to rigorously and formally prove that such set is not open in the product topology. Furthermore, is it necessary to us reductio ad absurdum to prove it?

$\endgroup$
2
$\begingroup$

You can directly show its complement is not closed. Let $x_n = (0,0,\dots,0,0,2,2,2,2,2,\dots)$ where there are $n$ $0$'s. Then each $x_n \in \mathbb{R}^\omega\setminus (-1,1)^\omega$ but $x_n \to (0,0,0,\dots) \in (-1,1)^\omega$.

$\endgroup$
1
$\begingroup$

In an infinite product like $\mathbb{R}^\omega$, every non-empty basic open subset $O$ has the property of "full projection":

$$\exists n \in \omega: \pi_n[O] = \mathbb{R}\tag{1}$$

As all non-empty open subsets are unions of non-empty basic open ones, and if $O$ has property (1), so has any superset of $O$, all non-empty open subsets of $\mathbb{R}^\omega$ have property (1).

But $\pi_n[(-1,1)^\omega] = (-1,1) \neq \mathbb{R}$ for all $n$, so that set cannot be open.

$\endgroup$
  • $\begingroup$ you also gave a proof that the empty set is not open $\endgroup$ – mathworker21 Dec 19 '18 at 9:24
  • $\begingroup$ @mathworker21 OK, I'll add that detail. $\endgroup$ – Henno Brandsma Dec 19 '18 at 9:25
0
$\begingroup$

It all depends on which definition of the product topology you would like to use. One proof I like is as follows:

The sets of the form $\prod_{i \in \Bbb N} X_i$, where each $X_i$ is open and $X_i = \Bbb R$ for all but finitely many $i \in \Bbb N$, form a basis of the product topology on $\Bbb R^\omega$. However, the set $S = (-1,1)^\omega$ contains no such subset. In particular, we note that $\pi_i(S)$ (the image of the $i$th projection map, applied to $S$) fails to contain $\Bbb R$ for any $i \in \Bbb N$. Since $S$ contains no basis element, $S$ cannot be open.

$\endgroup$
  • $\begingroup$ The set $S = \emptyset$ contains no such subset. In particular, we note that $\pi_i(S)$ fails to contain $\mathbb{R}$ for any $i \in \mathbb{N}$. Since $S$ contains no basis element, $S$ cannot be open. :)))))) $\endgroup$ – mathworker21 Dec 19 '18 at 2:31
  • $\begingroup$ @mathworker21 if we take $X_1 = \emptyset$ and $X_j = \mathbb{R}$ for $j > 1$ then $\prod X_i = \emptyset$, so the empty set is of that form. The projection property does only apply to nonempty sets, as you noted. $\endgroup$ – Carl Mummert Dec 19 '18 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.