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I'm being asked to see whether

$$ \prod_{i \in I}\overline{X_i} = \overline{\prod_{i \in I}X_i} $$

where $\prod_i X_i$ has the product or box topology. My argument seems to prove the equality in both cases, which I'm skeptical about. One inclusion always works: since $\overline{\prod_{i \in I}X_i} \subseteq \prod_{i \in I}\overline{X_i}$ occurs if and only if $\pi_i(\overline{\prod_{i \in I}X_i}) \subseteq \overline{X_i}$ for each projection $\pi_i$, it suffices to note that since these are continuous (in both topologies),

$$ \pi_i(\overline{\prod_{i \in I}X_i}) \subseteq \overline{\pi_i(\prod_{i \in I}X_i)} = \overline{X_i} \quad (\forall i \in I). $$

As for the other inclusion, here's where I suspect I'm missing some details: if $x \in \prod_{i \in I}\overline{X_i}$ and $U$ is an (basic) open set that contains $x$, then $U \cap \prod_{i \in I}X_i \neq \emptyset$ if and only if $\pi_i(U\cap \prod_{i \in I}X_i)$ is non empty for all $i$. Now, because

$$ \pi_i(U\cap \prod_{i \in I}X_i) = \pi_i(U) \cap \pi_i( \prod_{i \in I}X_i) = \pi_i(U) \cap X_i $$

and $\pi_i$ is open, since $x_i \in \pi_i(U)$ and $x_i \in \overline{X_i}$, we have that $\pi_i(U) \cap X_i \neq \emptyset$. Thus each (basic) open set that contains $x$ intersects the product, hence it is in the closure of the latter.

Can you spot any mistakes? I feel like this exercise was designed to show a demarcation between the product and box topologies, and so I was expecting some inclusion to fail in one of the cases.

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  • $\begingroup$ It is the same for box and product topologies. $\endgroup$ – William Elliot Sep 13 '18 at 8:45
  • $\begingroup$ You have to use the fact that basic open sets are products of opens too. See my proof. $\endgroup$ – Henno Brandsma Sep 13 '18 at 21:41
  • $\begingroup$ Am I not using that when saying that projections are open and so each coordinate is in the closure of each space? (i.e. that $\pi_i(U) \cap X_i$ is not empty). I do agree that I should have changed the name of the subspaces and given a family of ambient spaces as context. $\endgroup$ – Guido Sep 13 '18 at 22:23
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The equality does hold for both topologies. This is no problem, as this only shows the closure for some special subsets is the same. For general subsets closures are very different: the set of bounded sequences in $\mathbb{R}^{\mathbb{N}}$ is closed in the box topology on this product but dense in the normal product topology e.g.

Notationwise it would be better to have the $X_i, i \in I$ be the total spaces and $A_i \subseteq X_i$ the component subsets. Then I'd formulate the identity as

$$ \overline{\prod_{i \in I} A_i} = \prod_{i \in I} \overline{A_i}$$

in either topology. One other way to start is to note that $\prod_{i \in I} \overline{A_i}$ is closed as it equals $\bigcap_{i \in I} \pi^{-1}_i[\overline{A_i}]$ which is an intersection of closed sets (as projections are continuous) and hence closed. As clearly $\prod_{i \in I} A_i \subseteq \prod_{i \in I} \overline{A_i}$ the closure of the left hand side is a subset of the right hand side. If OTOH $(q_i)_i \in \prod_{i \in I} \overline{A_i}$, let $O:= \prod_{i \in I} O_i$ be a basic open neighbourhood of $(q_i)_i$; all we need is that $q_i \in O_i$ for all $i$ and $O_i$ is open in $X_i$ (this holds in both topologies (!)). Then for each $i$, $O_ i\cap A_i \neq \emptyset$ as $q_i \in \overline{A_i}$ and the intersection points together witness that $O \cap \prod_{i \in I} A_i \neq \emptyset$, so that $(q_i)_i \in \overline{\prod_{i \in I} A_i}$ as required.

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