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Is there an elementary example of the following:

$X$ is a topological space, and $p \in X$. There exists a neighborhood basis for $X$ at $p$, but there exists no countable neighborhood basis for $X$ at $p$.

I am having trouble coming up with an example where it is not possible to "sift out" a countable subset from the neighborhood basis which is itself a neighborhood basis, probably because my intuition is too attached to spaces that are at least first countable.

I should perhaps mention that I am using the definition where neighborhood of $p$ means an open subset of $X$ containing $p$.

This is not a homework problem; just something I started thinking about while reading chapter 2 of Lee's Introduction to Topological Manifolds.

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    $\begingroup$ Hello. Does this answer your question? Particularly, "there exists a neighbourhood basis for $X$ at $p$" is always true, because you can just take the set of all neighbourhoods. If you want something closer to the "balls of radius $1/n$" picture, you might ask for a neighbourhood basis which is totally ordered by inclusion - then you have to do a little bit of order theory to get an example, because you have to find an order type of uncountable cofinality. These are really unintuitive! The $\omega_1$-related examples at the link do the job. $\endgroup$ Feb 24 at 18:26
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    $\begingroup$ @IzaakvanDongen As is probably evident from my question, I did not originally appreciate that "there exists a neighborhood basis for $X$ at $p$" is always true, so thank you for that important comment. It seems obvious in hindsight of course, and since the question then reduces to "is there a topological space that is not first countable" it is possible to find many examples. In fact the book presents (at least) one, in an exercise. But I think the example given in the accepted answer is easier to picture than the one given in the book (which is a space of infinite sequences). $\endgroup$
    – ummg
    Feb 25 at 11:54

2 Answers 2

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Consider the finite complement (a.k.a. cofinite) topology on $\mathbb{R}$ that consists of $\emptyset$ along with subsets $A \subseteq \mathbb{R}$ such that $\mathbb{R} - A$ is finite.

Take any point $x \in \mathbb{R}$; we show that there is no countable neighborhood basis of $x$. Suppose for the sake of contradiction that we did have a countable neighborhood basis $\{B_n\}_{n \in \mathbb{Z}^+}$ of $x$. Consider the collection $\bigcup_{n \in \mathbb{Z}^+} (\mathbb{R} - B_n)$; since it is a countable union of finite sets, it is countable. Then there must exist some $y \in \mathbb{R}$ different from $x$ that is not contained in this union. By deMorgan's Law,

$$y \in \mathbb{R} - \bigcup_{n \in \mathbb{Z}^+} (\mathbb{R} - B_n) = \bigcap_{n \in \mathbb{Z}^+} B_n$$

The neighborhood $\mathbb{R} - \{y\}$ of $x$ does not contain any element of $\{B_n\}_{n \in \mathbb{Z}^+}$, by virtue of the fact that each $B_n$ contains $y$ (as we've just shown). This contradicts the hypothesis that $\{B_n\}_{n \in \mathbb{Z}^+}$ is a neighborhood basis of $x$.

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    $\begingroup$ Great example, thank you! $\endgroup$
    – ummg
    Feb 24 at 18:48
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Let $X$ be any uncountable set, and consider the so-called co-countable topology defined as follows: $$ \tau=\lbrace V\subset X \backslash C_{X}V \textit{is countable}\rbrace \cup \lbrace \phi\rbrace $$ you can verify that this family defines indeed a topology. let $p\in x$ and let $V_{n})_{n\geq 1}$ be a countable family of nbds of $p$.define $W=C_{X}(A\cup \lbrace q\rbrace $,where $A=\cup_{n\geq 1} C_X V_{n}$ (which is countable) and $q$ is any point in $X\setminus A\cup \lbrace p\rbrace $.if there is some $n$ such that $V_{n} \subset W$,then $A\cup \lbrace q\rbrace \subset C_X V_n$,in particular $q\in A$ which is clearly imposible by the choice of $q$.

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    $\begingroup$ I object to the use of $\phi$ for $\varnothing$. $\endgroup$
    – GEdgar
    Feb 24 at 18:11
  • $\begingroup$ What does $C_X$ mean? $\endgroup$
    – ummg
    Feb 24 at 18:46
  • $\begingroup$ Oh, complement in $X$ I suppose. $\endgroup$
    – ummg
    Feb 24 at 18:47
  • $\begingroup$ @ummg yes you are right $\endgroup$ Feb 24 at 18:49

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