New theorems with pi-character

[JSMassmann]

• Almost discrete + separable => countable.
• $T_1$ + well-based + has countable $\pi$-character => first countable.
• Has a cofinite topology + has countable $\pi$-character => Countable.

As recommanded in #1712. Any feedback?

[felixpernegger]

$T_1$ + well-based + has countable $\pi$ -character => first countable.

T_1 can be generalized to R0

[felixpernegger]

I just realized that we have almost discrete + not Sequentially discrete + t1 => Has countable pi character.

Let $X$ be a almost discrete space, $p$ the special point. All points except $p$ are isolated and thus always have a local pi base.

Assume it is not sequentially discrete and t1. Then there exist some sequence, $s_1, s_2, \dots $ which is injective (for this we need t1) and converges to $p$. WLOG $p$ isnt under the $s_i$. Then the $s_i$ singletons form a local pi base for $p$.

[felixpernegger]

This PR still leaves 46+ spaces unknown. I really have the feeling we are missing some implications

[prabau]

I just realized that we have almost discrete + not Sequentially discrete + t1 => Has countable pi character.

Let X be a almost discrete space, p the special point. All points except p are isolated and thus always have a local pi base.

Assume it is not sequentially discrete and t1. Then there exist some sequence, $s_1, s_2, \dots $ which is injective (for this we need t1) and converges to p . WLOG p isnt under the s i . Then the s i singletons form a local pi base for p .

Nice. It allows to derive that S131 has countable pi-character.

I don't think we even need T1. Suppose a sequence $(s_n)$ converges in $X$ and is not eventually constant. Necessarily its limit is $p$. By taking a subsequence, we can assume all $s_n\ne p$. Then the collection of singletons $\{s_n\}$ is a local pi-base for $p$, because ... (works the same even if the sequence is not injective).

[prabau]

Similar result: almost discrete + countably tight => countable pi-character.

proof: the special point $p$ is not isolated. So it belongs to the closure of $X\setminus\{p\}$. If $D$ is a countable subset of $X\setminus\{p\}$ with $p\in\overline D$, the singletons $\{x\}$ with $x\in D$ form a countable local pi-base for $p$.

This allows to derive that three more spaces have countable character: S131, S23 (Arens-Fort space), S96 (Appert space).

It is more general than #1807 (comment) since, as far as I can tell, almost discrete + ~ sequentially discrete => countably tight (not derivable right now from pi-base, but easy to see from the ideas of the previous result).
Should this be added too? Possibly. But it does not add anything for pi-character.

[prabau]

T908 almost discrete + separable => countable: One can replace separable with the more general ccc, with an even easier proof.

[prabau]

The last paragraph is irrelevant (and X is homogeneous anyway). Remove.

[JSMassmann]

Yeah, I realized that.

[prabau]

@JSMassmann I changed the title so later on, when scanning the list of PRs, we can have an idea of what this was about.

[prabau]

T909: I don't think the sentence starting with "By P135 ..." is correct. What did you have in mind?

When trying to use the Kolmogorov quotient as here, we usually say something like "Assume first that X is {T1}" at the top. then the argument as before. And then a separate paragraph at the end with something like: "The result with {R0} is obtained by passing to the Kolmogorov quotient."
No need to give more details, but you can convince yourself that all the properties involved in the theorems have the appropriate meta-properties and that everything works out as expected based on these meta-properties.

[prabau]

@JSMassmann On second thought, we need almost discrete + ~ sequentially discrete => countably tight to be able to derive @felixpernegger 's almost discrete + ~ sequentially discrete => has countable pi-character.
What do you think?

[JSMassmann]

T909: I don't think the sentence starting with "By P135 ..." is correct. What did you have in mind?

When @felixpernegger said you could replace $T_1$ with $R_0$, after briefly thinking about it, it made sense to me that you could do it with only $R_0$. But after looking at it more carefully, I understand that he just meant an appeal to metaproperties.

What do you think?

Sure, I can add that to this PR.

[JSMassmann]

Actually, I don't see how to prove it because I don't see how to connect the sequence $(s_n: n \in \mathbb{N})$ converging to $p$ with the set $A$ with $p \in \overline{A}$.

[prabau]

T909: I don't think the sentence starting with "By P135 ..." is correct. What did you have in mind?

When @felixpernegger said you could replace T 1 with R 0 , after briefly thinking about it, it made sense to me that you could do it with only R 0 . But after looking at it more carefully, I understand that he just meant an appeal to metaproperties.

It's not that it can be done by just appealing to metaprops, but trying to use R0 directly does not work. The current proof chooses $a \in A \setminus \{x\}$. But even when assuming R0, $a$ is not necessarily topologically distinguishable from $x$. In that case, one cannot choose $U_A\in\mathscr U$ avoiding the point $a$.

[JSMassmann]

Yes. My mistake was thinking that $a$ and $x$ must be topologically distinguishable, because $a \in A \setminus \{x\}$ and $x \notin A \setminus \{x\}$, but $A \setminus \{x\}$ mightn't be open since $x$ might not be a closed point.

[prabau]

Actually, I don't see how to prove it because I don't see how to connect the sequence ( s n : n ∈ N ) converging to p with the set A with p ∈ A ― .

Yeah, there may be a problem here. Need to think more. Maybe the result is not true and there would be a counterexample.

[prabau]

@JSMassmann @felixpernegger almost discrete + ~ sequentially discrete => countably tight is not true after all.
Here is a counterexample, with similar idea as the construction in #1712 (comment). Please check me on this.

Let Y = Fortissimo space (for example S155 of cardinality $\aleph_1$) with $p$ = non-isolated point.
Let Z = Sierpinski space $\{a,b\}$ with $b$ as isolated point and $a$ as non-isolated (closed) point.
And take X = quotient of disjoint union $Y\coprod Z$ with $p$ and $a$ identified. Call the identified point also $p$ for convenience.

• X is almost discrete
• X is not sequentially discrete, since it is not even T1: every nbhd of $p$ contains $b$. (or easier, Z embeds into X via the quotient map, and Z is not T1)
• X is not countably tight: Y embeds into X via the quotient map, and Y is not countably tight (it's a hereditary property)

[JSMassmann]

Ah, I see. That makes sense. I'll add almost discrete + ~sequentially discrete => has countable pi-character after all, then.

[prabau]

Suggested change

	By passing to Kolmogorov quotients and appealing to metaproperties, the theorem holds with only {P135}.
	This concludes the proof for the {P2} case.
	The general case with {P135} instead of {P2} follows by passing to the Kolmogorov quotient.

[prabau]

Suggested change

	Below is a proof that uses {P2} instead of {P135}:
	We first handle the case where $X$ is {P2}.

[prabau]

Suggested change

	Let $p$ be the point so that all points other than $p$ are isolated. Since $p$ is not isolated, $X \setminus \{p\}$ is dense so $p \in \overline{X \setminus \{p\}}$. Let $D \subseteq X \setminus \{p\}$ be countable with $p \in \overline{D}$. Then $\{\{x\}: x \in D\}$ forms a countable local $\pi$-base for $p$.
	Let $p$ be the unique non-isolated point.
	Then $p \in \overline{X \setminus \{p\}}$.
	Let $D \subseteq X \setminus \{p\}$ be countable with $p \in \overline{D}$. Then $\{\{x\}: x \in D\}$ forms a countable local $\pi$-base for $p$.

no need for so much redundancy :-)

[prabau]

I'll try to write a separate PR for the example in #1807 (comment) satisfying
π-Base, Search for Almost discrete + ~ Sequentially discrete + ~ Countably tight