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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?

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up vote 10 down vote accepted

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 by P. van Emde Boas, every metric space has a minimal subbase for its topology.

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You’re right. How did you come up with it? – Brian M. Scott Nov 25 '13 at 22:43
To be honest, I don’t think that I’ve ever seen one before. I was actually looking in the other direction: I fell asleep last night while starting to think about possible obstacles, if any, to showing that no subbase is minimal. – Brian M. Scott Nov 25 '13 at 23:07
Thank you for your response. I'm still wondering how $\mathcal S$ is a subbase since if we have an open interval $(a,b)$ such that $b\in\mathbb{R}$ with $(a,b)\cap\mathbb{Z}\neq\emptyset$, then in order to cover $\lfloor b\rfloor$ we will always need an $S_n\in\mathcal S$ such that $S_n$ contains $\lfloor b\rfloor$. But then we will always have $(\lfloor b\rfloor+1,\lfloor b\rfloor+2)\cup(a,b)$. Hence we can never generate $(a,b)$. Am I missing something here? – tcmtan Nov 26 '13 at 23:16
Hmmm, that's odd. I'm not sure why that's the case though, sorry :( I have an idea, what about try and copy and paste it on the 'Ask Question' page and edit it there just for the sake of seeing the preview. – tcmtan Nov 27 '13 at 1:09
@bof This is a clever construction. I was trying to prove that there was no minimal subbase for days but couldn't get anywhere. How did your intuition lead you to concluding the opposite? I'd like to get an insight to your thought processes and how you were led to construct a counterexample? Maybe you can add this to your answer so that it will help demystify some reasons to your construction? Thanks! – tcmtan Nov 27 '13 at 4:05

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