# Elementary example where a neighborhood basis exists but no countable neighborhood basis exists?

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.

• 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. Feb 24 at 18:26
• @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).
– ummg
Feb 25 at 11:54

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 $$x$$.

• Great example, thank you!
– ummg
Feb 24 at 18:48

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 $$q$$.

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