alpha_i miscellaneous metaproperties

[prabau]

The $\alpha_i$ properties for $i=1,1.5,2,3$ are preserved by countable products.

There is a paper by Nogura that proves it for $i=1,2,3$ for Hausdorff spaces. I provided a proof in general in https://math.stackexchange.com/questions/5122756. (Need to add the case $\alpha_{1.5}$ to that mathse post, will do it right now).

This allows to assert the $\alpha_1$ trait for six more spaces.

Plus various other metaproperties.

[yhx-12243]

Suggested change

	- This property is preserved by finite products (see {{mathse:5122756}}).
	- This property is not preserved by finite products (see {{mathse:5122756}}).

[prabau]

Thanks. Mindless cut and past on my part.

[prabau]

I will undo the change for alpha_4. Preferable to not claim anything at this point, until maybe we add more spaces in the future that we can point to within pi-base.

[prabau]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case.
But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

[felixpernegger]

very cool

[Moniker1998]

Apart from that, it is probably the case that the result is also true for countably infinite products. Again, see the mathse post and the cited references [Nogura1, thm 2.1] and [Nogura2, thm 2.2], which provide a proof in the Hausdorff case. But the general case needs to be checked in detail. I have not had the time to do it yet.

We can either approve this with just finite products and change it to countable products in a later PR. Or study the argument in Nogura and see if it works as is. If changes are needed and it can be salvaged, one of us can write up something on mathse (as another answer to the above post, or a different question). If anyone wants to do it, feel free to do it and announce it here before writing it up. I will also look at it.

There's no pressure to approve this. Let's just go with try to check countably infinite case for now.

[felixpernegger]

@prabau I think this PR would be a good place to also add the other metaporpeties (disjoint union + kolmogorov quotient) to the alpha i properties as well (one direction for quotient might actually not hold for alpha 1 or something)
EDIT: Nevermind it is also iff for alpha_1

[prabau]

@felixpernegger https://math.stackexchange.com/questions/5123855/does-the-arkhangelskii-alpha-1-property-hold-for-x-if-it-holds-for-its-kol

I also convinced myself of that fact, but hope you (or anyone out there) can provide an answer. If no one does, I'll post one in a few days.

[yhx-12243]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

[prabau]

This is not a difficult fact: Just use the equivalent definition of disjoint 𝑆ₙ's, and lift the set from quotient.

Great. Feel free to post an answer.

[felixpernegger]

I posted an answer. I dont really love it but whatever. Should be ok hopefully

[prabau]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

[prabau]

I think I need to have another mathse post asking what people mean by the $\alpha_i$ when they work with sequences instead of countably infinite sets. In the literature, many papers are very sloppy about what is meant exactly, or write definitions that technically don't always make sense.

[felixpernegger]

@felixpernegger Thanks for posting an answer. Unfortunately, lots of little things there don't quite make sense. I already made some comments in the post itself. There may be some more to be said. I know things can be fixed.

Thanks, I applied your suggestions. About finiteness, note that the definition for $\alpha_1$ (and probably \alpha_i) stays the same if we replace "countably infinite" by "countable" everywhere.

Maybe this simplifies things if we chabge this in pibase. It should also resolve your problem below about sequences (by requiring that differences of images are cofinite etc)

[felixpernegger]

Or better add remark that this is equivalent, since proof for triats etc really need infinite

[prabau]

Yeah, it would be good to mention something about $\alpha_i$ in terms of sequences. But let me post this mathse question I have in mind first and let's see if some people out there can clear all that confusion.

[prabau]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). "
@felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

[felixpernegger]

"It's easy to see that the definition for α1 is equivalent if we replace "countably infinite" by "countable" everywhere (i.e. allow finiteness). " @felixpernegger Sorry but this is not easy at all. It actually makes very little sense to me. Again, let me write this other post, and then feel free to explain it in an answer to that. Will get to it by the end of today.

The notion of "set converging to point" we use can trivially be extended for arbitrary cardinility. In particular, a finite set converges to every points.
Obviously removing the finiteness condition everywhere not a weaker notion as the original ($S$ has the be infinite since $S_i \setminus S$ is finite). For the other direction:

Lets say we remove the infinite condition. Suppose finitely many of the $S_i$ are infinite. Then just take $S$ as the union of those.
Suppose infinite many $S_i$ are infinite, let $M \subseteq \mathbb{N}$ the indices of those. Since $|M|=|\mathbb{N}|$, we can apply $\alpha_1$ with $(S_i)_{i \in M}$. The $S$ we get from there then also works for $\mathbb{N}$ (i.e. all $S_i$ again)

[prabau]

You said replace countably infinite with countable everywhere.
But must the $S_i$ one starts with be infinite? And must the number of the $S_i$ also be infinite?

[felixpernegger]

You said replace countably infinite with countable everywhere. But must the S i one starts with be infinite? And must the number of the S i also be infinite?

both to no technically.

[prabau]

For https://math.stackexchange.com/questions/5123855 I have posted the answer I originally had in mind for showing the metaprop for the Kolmogorov quotient.
Still planning to write a separate post regarding the proper definition of $\alpha_i$ in terms of sequences.

[prabau]

@Moniker1998 I have posted a proof for the countably infinite product of $\alpha_i$ spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

[Moniker1998]

@Moniker1998 I have posted a proof for the countably infinite product of α i spaces.

Can you review those posts at your convenience:

• https://math.stackexchange.com/questions/5122756 (two answers, one for finite products, one for infinite)
• https://math.stackexchange.com/questions/5123855 (for Kolmogorov quotient)

@prabau Sorry I haven't been checking github as much recently. And I've been laying off topology a bit.
Reviewing this might be a good warm up, sure. I think I'll start with the quotient.

[Moniker1998]

@prabau I've looked at the proof for the quotients, yes your proof looks good to me. I've left a comment about it. Next I'll check out the proof for products, but it'll be a bit longer as it's more involved.

[prabau]

For the Kolmogorov quotient, the other answer https://math.stackexchange.com/a/5124226 made no sense to me. Even though he claims "it is easy to see ..." in the first paragraph, it's not well-defined and very dubious. (That's why I wrote my own answer.)

Do you agree it makes little sense, or it does make sense and you can explain it to me?

[Moniker1998]

@prabau I honestly don't want to read an answer that isn't fleshed out. As you said, if we are considering finite subsets as sequences then something went wrong.

I prefer your answer.

[prabau]

As you said, if we are considering finite subsets as sequences then something went wrong.

Exactly my point. Not worth bothering about.

[Moniker1998]

@prabau so right now it looks to me that you haven't proven products for $\alpha_{1.5}$.
Your preliminary 2 does not go well for that property. Let's see this in practice:

Partition $X$ into $X_1, ..., X_m$. For a fixed $k$, either $X_k\cap S_n$ is finite for cofinitely many $n$, and so we can just take the sum of where $X_k\cap S_n$ is infinite, or there is infinitely many $n$ for which $X_k\cap S_n$ is infinite, and so we would apply the $\alpha_i$ property. Now the finite vs. infinite part is kinda irrelevant here, don't worry.
The concerning part is that in order to have $\alpha_{1.5}$ property is that we need all of the subsets of $X_1\cap S_n, ..., X_m\cap S_n$ to be cofinite in those simultaneously. And this sounds impossible since we only know the subsets of $X_k\cap S_n$ are cofinite in it for infinitely many $n$, and not all $n$.

[prabau]

Is it for the Kolmogorov quotient metaproperty or for the countable product?
Let me take a look.

[Moniker1998]

@prabau countable product; for the meta-property you've only proved it for $\alpha_1$, no?

[prabau]

You are absolutely right. My argument does not work for $\alpha_{1.5}$. I'll see if I can find a way to fix it.

[felixpernegger]

@prabau can you also add a metaproperty saying "A space has this property iff all of its countable subspaces have this property." or something? That would simplify the discussion in #1619.

[prabau]

@prabau can you also add a metaproperty saying "A space has this property iff all of its countable subspaces have this property." or something? That would simplify the discussion in #1619.

Good idea. I'll add that.

[prabau]

I am writing a workaround right now for the $\alpha_{1.5}$ case.

Now added.