A canonical example for spaces that aren't $1^{\text{st}}$ countable

Question

To get the feeling for metric spaces (or $1^{\text{st}}$ countable spaces, in general), I always find that visualizing them as $\Bbb R^3$ gives a good intuition of what is going on (locally, of course).

However, a general topological space $\Bbb X$ need not be $1^{\text{st}}$ countable and I have hard time visualizing the behavior of the neighborhood base near any particular point $x\in\Bbb X$.

Is there any good canonical example of a topological space $\Bbb X$ that is NOT $1^{\text{st}}$ countable?
How should I visualize the neighborhood of a point $x$ in such a space?

Answer 1

Here are four examples that I find useful.

The first two are easy. Let $Y$ be an uncountable set and $p$ a point not in $Y$, and let $X=\{p\}\cup Y$. Points of $Y$ are isolated, and the open nbhds of $p$ are precisely the sets $X\setminus F$ such that $F$ is a finite subset of $Y$. In other words, the topology on $Y$ is discrete, and the topology at $p$ is the cofinite topology.

If you change the topology so that the open nbhds of $p$ are precisely the sets $X\setminus C$ such that $C$ is a countable subset of $Y$, you get the second example. If $|X|=\omega_1$, $p$ has a nested local base in the second example but not in the first.

A compact example in which every point has a nested local base is the ordinal space $$\omega_1+1=[0,\omega_1]$$ with the order topology; every point except $\omega_1$ has a countable local base, and the sets $(\alpha,\omega_1]$ for $\alpha<\omega_1$ form a nested local base at $\omega_1$. If you remove the set of countable limit ordinals from this space, you get the second example above.)

The fourth example starts with a free ultrafilter $p$ on $\Bbb N$. The space is $X=\{p\}\cup\Bbb N$. Points of $\Bbb N$ are isolated. A set $U\subseteq X$ is an open nbhd of $p$ if and only if $p\in U$ and $U\cap\Bbb N\in p$. For a much bigger example you can take $\beta\Bbb N$, the Čech-Stone compactification of $\Bbb N$; no point in $\beta\Bbb N\setminus\Bbb N$ has a countable local base. If $p\in\beta\Bbb N\setminus\Bbb N$, the subspace $\{p\}\cup\Bbb N$ of $\beta\Bbb N$ is precisely the space described at the beginning of this paragraph.

Answer 2

If $X$ is not first countable then it is not metrizable, so of course it is hard to visualize. The typical example in mind are weak topologies and/or locally convex topologies (i.e. topologies given by families of seminorms). After you get some experience with weak and weak$^*$ neighbourhoods (that, for instance, contain subspaces) you will get some familiarity with non-metrizable spaces.

I want to comment on your "intuition". In my view, it is a bad one. Because in general, balls of metric spaces are not compact, so thinking of metric spaces locally as $\mathbb R^n$ might be misleading. One tipically gets used to $\ell^1(\mathbb N)$, $\ell^2(\mathbb N)$, and $\ell^\infty(\mathbb N)$ as basic examples of normed spaces; these are infinite-dimensional and so balls are not compact.

Answer 3

Example (1). Let $S$ be an uncountable set and take $p\not \in S.$ Define a topology $T$ on the set $S\cup \{p\}$ by : (i). Every subset of $S$ belongs to $T$. (ii). If $p\in A\subset S$ then $A \in T$ iff $S\backslash A$ is finite.

If $F$ is a countable family of open nbhds of $p,$ then $S\backslash \cup F$ is uncountable, so take any $q\in S\backslash \cup F.$ Then $V=(S\cup \{p\})\backslash \{q\}$ is a nbhd of $p$ and no member of $F$ is a subset of $V.$

Remark: $S$ is a discrete subspace of $S\cup \{p\}.$ The space $S\cup \{p\}$ with the topology $T$ is called the $1$-point compactification of the discrete space $S.$

Example (2). In example (1), change the phrase "iff $S\backslash A$ is finite" in condition (ii) of the def'n of T to "iff $S\backslash A$ is countable".

Example (3). If you are familiar with ordinals: The $\epsilon$-order topology on the ordinal $\omega_1+1$ is not first-countable. If $U$ is a nbhd of the point $\omega_1$ then there exists $x_U<\omega_1$ such that the interval $(x_U,\omega_1]$ is a subset of $U.$ So if $\{U(n) :n\in N\}$ is a family of nbhds of $\omega_1,$ let $y=\sup_{n\in N}x_{U(n)}.$ Then the interval $V=(y+1,\omega_1]$ is a nbhd of $\omega_1$ and no $U(n)$ is a subset of $V.$

Remark : There are many other types of examples. Even a countable space can have uncountable character.