Submetrizable implies Dieudonne complete (for completely regular spaces)

[Moniker1998]

No description provided.

[prabau]

the theorem number conflicts with #1819.

[Moniker1998]

@prabau do you want me to rename it

[prabau]

yes, please.

[Moniker1998]

@prabau comments on the substance?

[prabau]

I won't have time to look at it today. Hopefully tomorrow.

[Moniker1998]

@prabau this result is easier though, so maybe you want to review it first. Also, possibly more for you to do here, although I don't know if there's any other references for this.

[Moniker1998]

https://topology.pi-base.org/theorems/T000742 is essentially a version of this for realcompact spaces

Here we similarly have hereditary Dieudonne completeness, though we'll probably not add hereditary property on pi-base

The conclusion of hereditary Dieudonne completeness is essentially a stronger version of T742, given the two properties are equivalent for sizes < measurable.

Note this cannot be concluded in the same way as in T742. There is a T_4 metacompact space which is not Dieudonne complete. T382 does not translate to Dieudonne completeness

As far as I can see, all theorems that we have on pi-base about Dieudonne completeness and realcompactness are "strongest" versions. In the sense that when one can conclude stronger realcompactness, it is, and when weaker assumption of Dieudonne completeness can be used, it also is. The only case when this is not the case is when those are just non-theorems.

[Moniker1998]

I've added remarks on how realcompact and Dieudonne complete connect on the relevant pages. I think that's important, and I didn't do that before

[prabau]

Yes, I had noticed that for Hausdorff spaces with size < measurable, Dieudonne complete and realcompact are equivalent. So for space with such cardinality, T742 is actually an equivalent result (since the hypotheses are hereditary properties). I don't think we'll be having examples with higher cardinality any time soon. But is my understanding correct then that for higher cardinality one could have a T2 space that is Dieudonne complete and not realcompact?

[Moniker1998]

@prabau yeah, take a large enough metrizable space. Metrizable spaces are Dieudonne complete, and realcompact iff < measurable.

Note this is a corollary of a result which says that a $T_0$ Dieudonne complete space is realcompact iff every closed discrete subspace is < measurable (Shirota's theorem)

Even though in essence it won't add anything, I'd still like pi-base to have the sharpest theorems though

[prabau]

That reminds me of #1814. You were not in favor of adding such a space and maybe you are right. But maybe it could be useful, if we make clear what extra set-theoretic assumptions this would depend on. Just leaving this as a comment.

@StevenClontz FYI

[Moniker1998]

@prabau well I've already responded to that

[Moniker1998]

@prabau @felixpernegger can one of you review this

[felixpernegger]

This is more about P221 than about P162, so I would delete this sentence here

[prabau]

I think this sentence is useful and worth keeping. It basically says (implies) that for T0 spaces with reasonable cardinality, realcompact and Dieudonne complete are equivalent properties.

[Moniker1998]

@felixpernegger what do you mean?

[prabau]

@felixpernegger the main thing here is to verify the exercise 15U.3 in Gillman-Jerison. Unfortunately, I cannot say I understand it at this time. As it is not evident, it would be worthwhile to point to a detailed proof here. Were you able to verify the correctness of it yourself?

[Moniker1998]

@prabau what exactly do you want me to do here

[felixpernegger]

@prabau if you really want me to, but i think its fine.

The probability that 1) the book has an error at this exact spot and 2 @Moniker1998 did not notice it while solving the xercice is very low

[Moniker1998]

@felixpernegger @prabau I cannot vouch for its correctness but I found a proof in Topological groups and related structures by Arhangelskii and Tkachenko

[prabau]

@Moniker1998 Very nice find of theorem 6.10.8 in Arhangelskii-Tkachenko.

Hope you can confirm the following detail. In the first paragraph of the proof, it says that the topology $\mathscr T_f$ is submetrizable. But the second paragraph actually needs the $\mathscr T_f$ to be metrizable. The reason that's the case is that the topology $\gamma_f$ is pseudometrizable (with pseudometric $d(x,y)=|f(x)-f(y)|$) and the supremum (join) of a metrizable topology ($\mathscr T_0$) and a pseudometrizable topology ($\gamma_f$) is metrizable.

[Moniker1998]

@prabau yes

[prabau]

Looks good. Approving for now.
@felixpernegger Did you have any further issues?

[prabau]

I think this sentence is useful and worth keeping. It basically says (implies) that for T0 spaces with reasonable cardinality, realcompact and Dieudonne complete are equivalent properties.