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.
Best Answer
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.