Bases and pseudo-bases in topological spaces

Question

Let $X$ be a topological space and $x \in X$.

• A base for $X$ at $x$ is a collection $\mathcal{B}$ of open sets of $X$, all of them having $x$ as one of its elements such that for every open neighborhood $U$, of $x$, there is a $B\in \mathcal{B}$ such that $B \subseteq U$.
• A pseudo-base for $X$ at $x$, on the other hand, is a collection $\mathcal{B}$ of open sets, such that $\bigcap \mathcal{B}= \{x\}$.

For $T_1$ spaces, every base at a point is a pseudo-base at that same point.

However, I am having trouble coming up with examples of pseudo-bases that are not bases, so I would like some. Also, are there any sufficient conditions on $X$ that make these two notions coincide?

Answer 1

A not-too-enlightening example of a pseudo-base at a point which is not a local base at that point is the following. In the real line $\mathbb{R}$ consider the family $$ \mathcal{B} = \{ ( - \tfrac 1n , \tfrac 1 n ) \cup (n,n+1) : n \in \mathbb{R} \}. $$ $\mathcal{B}$ can be readily shown to be a pseudo-base at $0$, but it is clearly not a local base at $0$.

Similar to this, fix $V_{-1} = (1,2)$ and $V_{1} = (2,3)$, then $$\mathcal{B}^\prime = \{ \{ ( - \tfrac 1n , \tfrac 1 n ) \cup V_{(-1)^{n}} : n \in \mathbb{R} \}$$ is also a pseudo-base at $0$ which is not a local base at $0$.

These examples hint that the notions of (local) base and pseudo-base are difficult to make coincide for a point $x$ in a space $X$. Generalising the second example above, if there are two disjoint open sets $V , W \subset X$ such that $x \notin \overline{V} \cup \overline{W}$, then we can construct a pseudo-base at $x$ which is not a local base: Fix a local base $\mathcal{B}$ at $x$ and set $$\mathcal{B}^\prime = \{ U \cup V : U \in \mathcal{B} \} \cup \{ U \cup W : U \in \mathcal{B} \}.$$

The only T₁-spaces I can think of where every pseudo-base at a point is also a local base are the one-point space and the two-point discrete space.

As DanielWainfleets's answer indicates, the importance is generally on the "cardinal functions" defined in terms of these notions.

Answer 2

The character $\chi(x,X)$ of $x \in X$ is the least infinite cardinal $C$ such that there exists a base $B$ at $X$ whose cardinal is $\leq C.$ If $\chi(x,X)=\aleph_0$ for every $x\in X$ then $X$ is called first-countable.

If $X$ is $T_1$ then the pseudo-character $\psi(x,Z)$ of $x\in X$ is the least infinite cardinal $P$ such that $\{x\}=\cap F$ for some open family $F$ whose cardinal $\leq P.$

We always have $\psi(x,X)\leq \chi(x,X)$ for a $T_1$ space $X$.

If $X$ is a metric space then $\psi(x,Z)=\chi(x,X)=\aleph_0.$ If X is a linear space we have $\psi(x,X)=\chi(x,X).$ To get $\psi(x,X)=\aleph_0$ and $\chi(x,Z)\geq \aleph_1$ for some $x\in X$ we need something less ordinary.

Let $\Bbb R$ have the usual (standard) topology and let $X$ be the quotient space $\Bbb R_{/\Bbb N}.$ Equivalently, $X=(\Bbb R$ \ $\Bbb N) \cup \{p\}$ with $p\not \in \Bbb R$ where:

(i). If $p\in S\subset X $ then $S$ is open in $X$ iff $(S$ \ $\{p\})\cup \Bbb N$ is open in $\Bbb R.$

(ii). If $p\ne S\subset X$ then $S$ is open in $X$ iff $S$ is open in $\Bbb R.$

We have $\psi(p,Z)=\aleph_0$ because $S_n=\{p\}\cup (\cup_{m\in \Bbb N}(m-1/n,m)\cup (m,m+1/n)\;)$ is open in $X$ for every $n\in \Bbb N$ and we have $\cap_{n\in \Bbb N}S_n=\{p\}.$

Let $B=\{B_n:n\in \Bbb N\}$ be a countable open family in $X$ with $p\in B_n $ for every $n.$ For each $n$ there is a function $f_n:\Bbb N\to \Bbb R^+$ such that $(-f_n(m)+m,f_n(m)+m) \backslash \{m\} \subset B_n$ for every $m\in \Bbb N.$

Let $g(m)=\frac {1}{3}\min \{f_n(m): n\leq m\}$ for each $m\in \Bbb N.$ Let $A=\{p\}\cup (\cup_{m\in \Bbb N}(-g(m)+m,g(m)+m)\;).$ Then $A \backslash \Bbb N$ is open in $X$ and $ p\in A\backslash \Bbb N.$ But no $B_n$ is a subset of $A\backslash \Bbb N$ because $n+\frac {2}{3}f_n(n)\in B_n$ \ $(A \backslash \Bbb N).$

So no countable open family $B$ can be a base at $p$ in $X.$

BTW. With the usual topology on $\Bbb R$ the Box-Product topology on $\Bbb R^{\Bbb N}$ also has countable pseudo-character and uncountable character but I think this answer is already long enough

Answer 3

Plainly, a countable $\text T_1$ space has a countable pseudo-base at every point. This is not true with "pseudo-base" replaced by "base".

As in my answer to this old question, define a topology on $\mathbb N=\{1,2,3,\dots\}$ by calling a set $S\subseteq\mathbb N$ open if
$$\text{either}\quad1\notin S\quad\text{or else}\quad\sum_{n\in\mathbb N\setminus S}\frac1n\lt\infty.$$ With this topology $\mathbb N$ is a regular Hausdorff space, also zero-dimensional and normal, and of course it has a countable pseudo-base at every point. However, it is easy to see that there is no countable base for the neighborhoods of $1$.