Minimal subbases for the canonical topology on $\mathbb{R}$?

Question

Thoughts: I know that every base for the canonical topology can be reduced. So I'm guessing that maybe the canonical topology doesn't have a minimal subbase? Would it be possible to use the fact that every base for the canonical topology on $\mathbb{R}$ is not minimal? Or can we directly deal with subbases?

If we take the canonical topology on $\mathbb{R}$ then a subbase for the canonical topology would be the collection of sets $(-\infty,b)$ and $(a,\infty)$ where $a,b\in\mathbb{R}$. Call this collection $\Delta$. It is then easy to see that we can always take an element out because if we take out $(-\infty,x)$ for some $x\in\mathbb{R}$, then every $(a,x)$ for any $a\in\mathbb{R}$ is still a union of elements in the collection of all finite intersections of $\Delta\backslash\{(-\infty,x)\}$ . So inductively we see that subbases of the form above can always be reduced.

So my hunch is that in general there is no minimal subbase for the canonical topology on $\mathbb{R}$. But I'm not sure though. Is it possible to construct a minimal subbase?

Answer 1

As a matter of fact, it does have a minimal subbase.

Let $B_1,B_2,\dots$ enumerate the collection of all open intervals, with rational endpoints, of length less than $1$.

Define a bijection $f:\mathbb N\to\mathbb Z$ so that $B_n\cap[f(n),f(n)+2]=\emptyset$ for each $n\in\mathbb N$.

For each $n\in\mathbb N$ define a set $S_n\supset B_n$ as follows:

• if $B_n\cap\mathbb Z=\emptyset$, then $S_n=B_n\cup\left(f(n)-\frac12,f(n)+\frac12\right)$;
• if $B_n\cap\mathbb Z=\{m\}$, then $S_n=B_n\cup\left(f(n)-\frac12,f(n)+\frac12\right)\cup(m+1,m+2)$.

First, I will show that $\mathcal S$ is a subbase for $\tau$, the usual topology of $\mathbb R$. Clearly, the elements of $\mathcal S$ are open in $\tau$. Let $\mathcal S^*$ be the collection of all finite intersections of elements of $\mathcal S$. Let $x\in\mathbb R$ and $\varepsilon\gt0$ be given; I have to find a set $S\in\mathcal S^*$ such that $x\in S\subseteq(x-\varepsilon,x+\varepsilon)$. Choose $i,j\in\mathbb N,i\ne j,$ so that$$x\in B_i\subseteq(x-\varepsilon,x+\varepsilon),\ B_i\cap(\mathbb Z\setminus\{x\})=\emptyset,$$$$x\in B_j\subseteq(x-\varepsilon,x+\varepsilon),\ B_j\cap(\mathbb Z\setminus\{x\})=\emptyset.$$

Case I, $x\notin\mathbb Z$. Then $B_i\cap\mathbb Z=B_j\cap\mathbb Z=\emptyset$, so $S_i=B_i\cup\left(f(i)-\frac12,f(i)+\frac12\right)$ and $S_j=B_j\cup\left(f(j)-\frac12,f(j)+\frac12\right)$. Let $S=S_i\cap S_j\in\mathcal S^*$; then$$x\in B_i\cap B_j\subseteq S\subseteq B_i\cup B_j\subseteq(x-\varepsilon,x+\varepsilon).$$

Case II, $x\in\mathbb Z$. Then $B_i\cap\mathbb Z=B_j\cap\mathbb Z=\{x\}$, so$$S_i=B_i\cup\left(f(i)-\frac12,f(i)+\frac12\right)\cup(x+1,x+2),$$$$S_j=B_j\cup\left(f(j)-\frac12,f(j)+\frac12\right)\cup(x+1,x+2),$$$$x\in S_i\cap S_j\subseteq B_i\cup B_j\cup(x+1,x+2)\subseteq(x-\varepsilon,x+\varepsilon)\cup(x+1,x+2).$$Choose $k\in\mathbb N$ so that $f(k)=x$, and let $S=S_i\cap S_j\cap S_k\in\mathcal S^*$. Since $x\in S_k$ and $S_k\cap(x+1,x+2)=\emptyset$, we have $x\in S\subseteq(x-\varepsilon,x+\varepsilon)$.

Finally, I will show that $\mathcal S$ is a minimal subbase for $\tau$. Observe that, if $m\in\mathbb Z$ and $m\in S_n$, then either $m\in B_n$ (and so $(m+1,m+2)\subset S_n$), or else $m=f(n)$. Hence, if $\mathcal S'=\mathcal S\setminus\{S_n\}$ for some $n\in\mathbb N$, and if we set $m=f(n)$, then (recalling that $f$ is a bijection) every element of $\mathcal S'$ which contains the point $m$ also contains the interval $(m+1,m+2)$. Thus the topology generated by $\mathcal S'$ does not contain all of the usual neighborhoods of the point $m$.

P.S. According to this article, every metric space has a minimal subbase for its topology:
van Emde Boas, P. (1967). Minimality of subbases and bases of topological spaces. Stichting Mathematisch Centrum. Zuivere Wiskunde. Stichting Mathematisch Centrum.