Has closed discrete subset of size 𝔠 (part 4)

[yhx-12243]

Continuation of #1584, #1592 and #1597.

[Moniker1998]

Let's add a meta-property for P198 about unions.

[prabau]

P227: "If a closed subspace of $X$ satisfies this property, so does $X$."
That's a true fact. But I am not sure we need to add this as a meta-property. It does not quite fit into the mold of other meta-properties, and it's also kind of immediate. If we need that to justify some trait, we can just do it directly without referring to this as a meta-property.

(For comparison, consider the property "cardinality at least 4". It would be true to say: "If a subspace of $X$ has the property, so does $X$." But it's not very helpful, it's just too trivial.)

At least, let me review the rest to see how it is used.

[Moniker1998]

P227: "If a closed subspace of X satisfies this property, so does X ."
That's a true fact. But I am not sure we need to add this as a meta-property. It does not quite fit into the mold of other meta-properties, and it's also kind of immediate. If we need that to justify some trait, we can just do it directly without referring to this as a meta-property.

I disagree. We should collect meta-properties when we can.

(For comparison, consider the property "cardinality at least 4". It would be true to say: "If a subspace of X has the property, so does X ." But it's not very helpful, it's just too trivial.)

We can add it once we're using it in an argument.

[prabau]

Let's see what other people have to say.

[yhx-12243]

P227: "If a closed subspace of X satisfies this property, so does X ." That's a true fact. But I am not sure we need to add this as a meta-property. It does not quite fit into the mold of other meta-properties, and it's also kind of immediate. If we need that to justify some trait, we can just do it directly without referring to this as a meta-property.

(For comparison, consider the property "cardinality at least 4". It would be true to say: "If a subspace of X has the property, so does X ." But it's not very helpful, it's just too trivial.)

Basically, this is what I called “reverse hereditary” before.

And the former discussion about how to say “reverse hereditary” is exactly “If a closed subspace of X satisfies this property, so does X.”

[prabau]

Ok. If both of you think it's valuable, I won't object.

[prabau]

see #1242

[prabau]

General discussion about meta-properties related to "union".

In general, given a property P, several of the meta-props are phrased as "This property is preserved by such and such a construction". Typically we are given a topological space (or a collection of them) which has the property P, and out of that we construct another space, which then also has the property.

So for "union", what do you want to mean here by "preserved by union"? At first glance, it does not quite fit the above paragraph? Maybe it would be better to rephrase things in a different way.
Can you elaborate the intended concept?

[Moniker1998]

@prabau that if $X = \bigcup_i X_i$ and $X_i$ have P (as subspaces), then $X$ has P

I think this is clear, and we don't need to always talk about constructions. Just because something has seemed like a theme so far, doesn't mean it is.

[prabau]

Exactly! So let's phrase it that way, and not in terms of "constructions that preserve a property".

Something like this for example:
"If $X$ is covered by countably many subspaces, each having the property, then so does $X$."

[Moniker1998]

I am not confused by this, but if you think it's confusing, feel free to standarize your own version, here and in the wiki.

[prabau]

Talking about "preserved by union" is very confusing. One reason is that, apart from the topological disjoint union (= categorical coproduct) of spaces, there is also the notion of (non-disjoint) union, where one starts with a collection of spaces, not necessarily disjoint, and one constructs a new space that is the union of all the spaces as a set, with a certain topology that satisfies some universal property. That construction is not what is meant here. So "preserved by union" is a misleading choice.

[Moniker1998]

Talking about "preserved by union" is very confusing. One reason is that, apart from the topological disjoint union (= categorical coproduct) of spaces, there is also the notion of (non-disjoint) union, where one starts with a collection of spaces, not necessarily disjoint, and one constructs a new space that is the union of all the spaces as a set, with a certain topology that satisfies some universal property. That construction is not what is meant here. So "preserved by union" is a misleading choice.

No because that construction has a name. I don't see a reason to continue this discussion though, we've already agreed that we can do what you prefer.

[felixpernegger]

I am not confused by this, but if you think it's confusing, feel free to standarize your own version, here and in the wiki.

I also think it is unclear what is meant by This property is preserved by countable unions.

"If X is covered by countably many subspaces, each having the property, then so does X ."

this sounds good to me

[prabau]

S79-P198: I added a commit as a suggestion was not possible. Please check.

[prabau]

S81 (Alexandroff plank): Did not spend time thinking about it, but can you explain to me here why each of the subspaces has countable extent?

[yhx-12243]

S81 (Alexandroff plank): Did not spend time thinking about it, but can you explain to me here why each of the subspaces has countable extent?

The former two product is Compact × Lindelöf ⇒ Lindelöf ⇒ Countable extent. The third one is countably compact.

[prabau]

S81 (Alexandroff plank): Did not spend time thinking about it, but can you explain to me here why each of the subspaces has countable extent?

The former two product is Compact × Lindelöf ⇒ Lindelöf ⇒ Countable extent. The third one is countably compact.

Great. I think it would be good to add a few words about that, maybe reorganize as a bullet list with explanation for each as needed. Or some other way to present things if you prefer.

[prabau]

It all looks good to me. Leaving it to @Moniker1998 to approve, in case he has further comments.

[Moniker1998]

We could add reference to product of compact and Lindelof space is Lindelof if we're going to use this argument. This isn't completely trivial.

[prabau]

good idea

[Moniker1998]

@prabau do we keep this reference?

[prabau]

I think the current reference is ok.

[Moniker1998]

@prabau I'm not sure, the current reference is something like "I have this proof and can someone check my work?". And so it seems a bit... you know.

[prabau]

It is not very good as a mathse question, and it's way too verbose. But it has the advantage of being pretty direct. Your post for example is much more informative, but not as accessible maybe.

For another possibility, p. 576 of Handbook of set-theoretic topology ?

[Moniker1998]

Your post for example is much more informative, but not as accessible maybe.

I think it is accessible once you understand the concepts within. And they are not hard. It should be accessible to anyone willing to give it a try. It might be intimidating at a first glance, but it isn't.

[Moniker1998]

I'm not saying we should use it as a reference, just saying it only seems inaccessible.

[prabau]

I think it is accessible once you understand the concepts within. And they are not hard. It should be accessible to anyone willing to give it a try. It might be intimidating at a first glance, but it isn't.

I totally agree that once one reads it carefully it makes total sense, and it provides more information too. Just that it requires more notions to get to the result (compact x Lindelof = Lindelof), so it makes that particular result less accessible, i.e., requiring more notions if a pi-base user is seeing this for the first time and just wants to get a proof of that result. In that sense, it may not be as appropriate as a pi-base justification in this case. Just my impression here.

[Moniker1998]

We should use meta-property instead, {S137|P227} and {S137} is a closed subspace of {S136}.

[prabau]

True, one could do that. In this case though, is it worth doing? S136 (Bing's example G) is certainly the main space and the result is immediate for this one. S137 (Michael's subspace) is a kind of "derivative space" from S136, it's less "obvious" in that sense than S136. And the result is also immediate for S137 (in both cases because $M$ belongs to both spaces). So it does not seem worth using a meta-property here; or at least, for S136, the meta-property does not provide any easier justification in this case.

[Moniker1998]

@prabau It's in line with our practices. It doesn't matter if it's worth adding, it should be added.

[prabau]

Not sure about that. We have used meta-properties in many places because it makes the justifications simpler. In this case, it's not clear it simplifies anything. But if you are adamant about it, I'll leave it to @yhx-12243 to decide.

[prabau]

(Sorry, but thinking about it some more, we use meta-properties in general to prove a trait by deferring to an easier/simpler space in general. Here, S136 is the simpler space in a way, in the sense that it is the one that is first defined; S137 is a subspace of it that is not defined in isolation from it. So it makes little sense to me to defer to a trait of the subspace in this case. The traits of S136 can be decided in a self-contained way without involving S137. But again, I'll let @yhx-12243 decide.)

[prabau]

@Moniker1998 isn't using the meta-property more work here, as one needs to verify (in one's mind) that S137 is a closed subspace of $X$, which is not needed if one just talks about $M$ ?

[Moniker1998]

@prabau That's in the description of S137, it's the whole point of this space, that it's a closed subspace of S136.

[Moniker1998]

Sorry, but thinking about it some more, we use meta-properties in general to prove a trait by deferring to an easier/simpler space in general. Here, S136 is the simpler space in a way, in the sense that it is the one that is first defined; S137 is a subspace of it that is not defined in isolation from it. So it makes little sense to me to defer to a trait of the subspace in this case. The traits of S136 can be decided in a self-contained way without involving S137. But again, I'll let @yhx-12243 decide

This is just false, we don't do that just when the subspace is simpler.

[Moniker1998]

@yhx-12243 I'd move the argument that {S137|P227} to the front, and then say "explicitly, ...". Then it will flow more naturally.

[prabau]

That's fine too.

[Moniker1998]

@prabau of course I'm not approving it, my comments haven't been taken into consideration and changes weren't made

[prabau]

This seems a proof choice of reference. Even the title "... compact set ..." is bad.
A better one could be https://math.stackexchange.com/questions/3652735
but we should probably shorten the overly long title to "If X is Lindelof and Y is compact, then X×Y is Lindelof".

Have you checked in Engelking or Munkres directly?

[Moniker1998]

Yes the math.se link is a non-answer. There's nothing there.

I've posted this https://math.stackexchange.com/questions/477142/if-x-and-y-are-lindel%c3%b6fs-respectively-countably-compact-spaces-so-that-x/5121472#5121472 with a proof but I didn't do it for pi-base, just to prove it for myself. But probably better to use a book as a reference.

[yhx-12243]

This seems a proof choice of reference. Even the title "... compact set ..." is bad. A better one could be https://math.stackexchange.com/questions/3652735 but we should probably shorten the overly long title to "If X is Lindelof and Y is compact, then X×Y is Lindelof".

Have you checked in Engelking or Munkres directly?

Engelking: There are extra separation assumptions.
Munkres: Only exercise.

[prabau]

I've posted this https://math.stackexchange.com/questions/477142/if-x-and-y-are-lindel%c3%b6fs-respectively-countably-compact-spaces-so-that-x/5121472#5121472 with a proof but I didn't do it for pi-base, just to prove it for myself. But probably better to use a book as a reference.

@Moniker1998 Thanks for the interesting post. I need to read it in detail.

Apart from that, the mathse ref that @yhx-12243 just added seems good enough and complete, even if it's too wordy. It's basically the tube lemma.

[Moniker1998]

@prabau sure, I've added more to it to make more parallels and so on, make it less of a mystery.