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Initially, I was looking for an example of a $T_1$ space that has a countable network and that doesn't have countable pseudocharacter. I was able to determine that the cofinite topology on $\mathbb{R}$ is an example of such. However, I am still wondering if anyone knows an example of a $T_1$ space that has a countable $k$-network and that doesn't have countable pseudocharacter. I haven't been able to determine or find any yet. One possible candidate could be: Real numbers extended by a point with co-countable open neighborhoods.

Almanzoris
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    I think it's better to add a direct link to the space you're referring to. – Jakobian Sep 16 '24 at 19:49
  • Alright, was just to show that it is the only pending one in $\pi$-Base. – Almanzoris Sep 16 '24 at 19:51
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    Do you know if $T_2$ + countable network implies countable pseudocharacter? – Jakobian Sep 16 '24 at 20:28
  • I was wondering about it too, but wanted ask this question first. I don't know if $T_2$ + countable network implies countable pseudocharacter yet. – Almanzoris Sep 16 '24 at 20:50
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    Let $X$ be a $T_2$ and hereditarily Lindelöf space. Given a point $x$ of $X$. $X \setminus {x}$ is Lindelöf, and for each $y \in X \setminus {x}$, there exists an open neighbourhood $A_y$ of $y$ in $X$ such that $x$ is not in the closure of $A_y$. A is an open cover of $X \setminus {x}$ in $X$. So it has a countable subcover. And $X \setminus {x}$ would be equal to the union of the closures of the elements of the subcover. So, it would be $F_{\sigma}$. Therefore, ${x}$ is a $G_{\delta}$ set. – Almanzoris Sep 16 '24 at 21:00
  • Yeah, it is sufficient to be $T_2$. – Almanzoris Sep 16 '24 at 21:01
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    Would be nice to have this on pi-base too. – Jakobian Sep 16 '24 at 21:05
  • Typo: the open cover is ${A_y}_{y\in X\setminus{x}}$. – Almanzoris Sep 16 '24 at 21:07
  • Yeah, I will add it too. – Almanzoris Sep 16 '24 at 21:07
  • Would you like to ask the question, so that we can cite it in the proof? – Almanzoris Sep 16 '24 at 21:09
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    No, I don't take credit for this. If anything you can write question and answer about it. Just check the box below which says you want to answer your question when posting it. This way you will post both a question and answer to it at the same time. Such practice is actually encouraged here (I mean, sure some people might think you shouldn't do that, but the site itself encourages that). – Jakobian Sep 16 '24 at 21:11
  • Alright, I will. Thank you for this information. Still, I had to ask. – Almanzoris Sep 16 '24 at 21:17

1 Answers1

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In this answer it's shown that compact subsets of $X = \mathbb{R}\cup\{\infty\}$ with $\mathbb{R}$ having Euclidean topology and open neighbourhoods of $\infty$ being of the form $U\cup \{\infty\}$ where $U\subseteq \mathbb{R}$ is open and co-countable are precisely of those subsets $K\subseteq X$ such that $K\cap\mathbb{R}$ is compact. This space is $T_1$ and doesn't have countable pseudocharacter.

Let $\mathcal{N}$ be a countable pseudobase for $\mathbb{R}$, then if $K\subseteq U$ where $K$ is compact and $U$ is open, pick $N\in\mathcal{N}$ such that $K\cap\mathbb{R}\subseteq N\subseteq U$. If $K\subseteq \mathbb{R}$ we are done, if $\infty\in K$ then replace $N$ by $N\cup\{\infty\}$.

Hence sets of the form $N$ or $N\cup\{\infty\}$ for $N\in\mathcal{N}$ form a countable pseudobase for $X$ so that $X$ has countable $k$-network.

Jakobian
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