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According to Munkres' Topology:

Definition. A space $X$ is said to have a countable basis at $x$ if there is a countable collection $\mathscr B$ of neighborhoods of $x$ such that each neighborhood of $x$ contains at least one of the elements of $\mathscr B$. A space that has a countable basis at each of its points is said to be first-countable.

Considering this, I guess that any space $X$ is first-countable. If there is any space $X$ that is not first-countable, please mention it with a simple explanation so that I understand better this concept.

Thanks a lot.

Silvia Ghinassi
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2 Answers2

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Let $\tau=\{\varnothing\}\cup\{\Bbb R\setminus F:F\text{ is finite}\}$; this is a topology on $\Bbb R$, called the cofinite topology. (A cofinite topology can be defined on any set.) Because $\Bbb R$ is uncountable, $\tau$ is not first countable.

To see this, let $x\in\Bbb R$, and suppose that $\mathscr{B}$ is a countable family of open nbhds of $x$, say $\mathscr{B}=\{B_n:n\in\Bbb N\}$. For each $n\in\Bbb N$ there is by definition a finite $F_n\subseteq\Bbb R$ such that $B_n=\Bbb R\setminus F_n$. Let $C=\{x\}\cup\bigcup_{n\in\Bbb N}F_n$; $C$ is the union of countably many finite sets, so $C$ is countable. $\Bbb R$ is uncountable, so we can choose a point $y\in\Bbb R\setminus C$. Let $U=\Bbb R\setminus\{y\}$. By definition $U$ is open, and clearly $x\in U$. However, for any $n\in\Bbb N$ we have $y\in B_n\setminus U$, so $B_n\nsubseteq U$, and therefore $\mathscr{B}$ is not a local base at $x$. Since $\mathscr{B}$ was any countable family of open nbhds of $x$, this shows that $\langle\Bbb R,\tau\rangle$ is not first countable at $x$. (And since $x$ was an arbitrary point of $\Bbb R$, we’ve actually shown that $\langle\Bbb R,\tau\rangle$ is nowhere first countable: no point has a countable local base.

Brian M. Scott
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There are lots of counterexamples:

  • The simplest is the co-countable topology on an uncountable set.

  • Slightly more complicated is the long line, or (what's essentially the same for our purposes) the ordinal $\omega_1+1$ with the order topology.

  • More naturally for many purposes, consider the space of all ultrafilters on $\mathbb{N}$.

A detailed explanation of the simplest example:

"Cocountable" just means "the complement is countable."

Fix some uncountable set $X$ - e.g., $X=\mathbb{R}$. The open sets in the cocountable topology are those which are (either the empty set or) all but countably much of $X$. For example, the irrationals form an open set in the cocountable topology on $\mathbb{R}$.

Now, fix a point $x\in X$, and suppose I had a countable neighborhood basis $\{N_i: i\in\mathbb{N}\}$ of $x$. It's a good exercise to show that, since each $N_i$ is cocountable, the intersection $N$ of all the $N_i$ is again cocountable. But now delete one point (other than $x$) from $N$ to get a smaller, but still cocountable, set $M$. Clearly $M\not\supseteq N_i$ for any $i$, but $M$ is an open neighborhood of $x$.

Noah Schweber
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