# Natural non-trivial topology on ${\mathbb R}$ such that there are more than $2^{\mathbb N}$ open sets

The collection of open sets of ${\mathbb R}$ with standard topology only has cardinality $2^{\mathbb N}$ because we can map the collection of countable sets of intervals with rational endpoints onto the collection of open sets. Of course if we declared all subsets of ${\mathbb R}$ to be open, then the cardinality of the collection of open sets would higher. But is there a "natural" (e.g. interesting folklore) example of a non-trivial topology on ${\mathbb{R}}$ such that the number of open sets is more than $2^{\mathbb N}$?

## 2 Answers

The Michael line $M$ is a classic example: it’s obtained from $\Bbb R$ by isolating each irrational and leaving each rational with its usual local base. That is, if $\tau$ is the usual topology on $\Bbb R$, $\tau\cup\big\{\{x\}:x\in\Bbb R\setminus\Bbb Q\big\}$ is a base for the topology of $M$. $M$ is hereditarily paracompact, and each finite power of $M$ is paracompact, but $M^\omega$ is not paracompact, and $M\times(\Bbb R\setminus\Bbb Q)$, where $\Bbb R\setminus\Bbb Q$ has its usual metric topology, is not even normal. Since $M$ has $2^\omega=\mathfrak{c}$ isolated points, it clearly has $2^{\mathfrak{c}}$ open sets.

Another is sometimes called the rational sequence topology. Briefly, it’s obtained by isolating each rational, assigning to each irrational $x$ a sequence $\langle q_n^x:n\in\omega\rangle$ of rationals converging to $x$ in the usual topology, and letting the sets $\{x\}\cup\{q_n^x:n\ge m\}$ for $m\in\omega$ be a local base at $x$. The resulting space is locally compact, Hausdorff, and zero-dimensional, but since it’s separable and has a closed, discrete subset — the irrationals — of cardinality $2^\omega$, it’s not normal. For each $A\subseteq\Bbb R\setminus\Bbb Q$ the set $A\cup\Bbb Q$ is open, so this space also has $2^\mathfrak{c}$ open sets.

The Sorgenfrey line is not an example: if $U$ is open in the Sorgenfrey line, there are a Euclidean open set $V$ and a countable set $C$ such that $U=V\cup C$, and since there are only $2^\omega$ Euclidean open sets and $2^\omega$ countable subsets of $\Bbb R$, there are only $2^\omega\cdot2^\omega=2^\omega$ Sorgenfrey open sets.

• Thanks! These are somewhat tautological examples as far as the cardinality argument goes for number of open sets, but it definitely answers my question as to whether there really are "real" examples of topologies that are studied because they have interesting properties, that also have the large number of open sets requested. – user2566092 Sep 26 '13 at 20:19
• @user2566092: You’re welcome! For an example in which the cardinality of the topology is a little less obvious, see Etienne’s answer; I’d forgotten about the density topology. – Brian M. Scott Sep 26 '13 at 20:24
• Given that GCH cannot be disproven, can we assert that any example (or at least any example we can constructively describe) must contain the powerset of some continuum sized set? – Ross Millikan Sep 26 '13 at 20:34
• @Ross: The rational sequence topology doesn’t, so far as I can tell. – Brian M. Scott Sep 26 '13 at 20:38
• The sense I was thinking of is that the rational sequence topology has the power set of $\Bbb {R \setminus Q}$, but I see that you have to pair each of them up with $\Bbb Q$ – Ross Millikan Sep 26 '13 at 20:43

A good candidate is the so-called density topology, which is actually quite useful in potential theory. By definition, a set $A\subset\mathbb R$ is open for this topology iff $A$ is Lebesgue-measurable and has density $1$ at each point $x\in A$, i.e. $$\forall x\in A\;:\; \lim_{h\to 0^+}\,\frac{m(A\cap [x-h,x+h])}{2h}=1\, ,$$ where $m$ is Lebesgue measure.

One property of this topology (call it $\tau$) is that all measurable subsets of $\mathbb R$ with Lebesgue measure $0$ are $\tau$-closed. So there are $2^{\mathbf c}$ closed sets wrt to $\tau$ (and hence $2^{\bf c}$ open sets).

If you are interested, you can have a look at the following paper and the references therein: http://msp.org/pjm/1976/62-1/pjm-v62-n1-p25-p.pdf

• Nice! This example is not only a "natural" topology worthy of study, but it's also not immediately obvious that the number of open sets is larger than $2^{\mathbb N}$. I was secretly hoping for an example where the cardinality of the collection of open sets was not trivial to show. – user2566092 Sep 26 '13 at 20:23
• One small detail...shouldn't the definition be that the limit is equal to $2$ instead of $1$, or alternatively that the interval is actually $[x - h/2, x + h/2]$? – user2566092 Sep 26 '13 at 20:33
• Yes, of course! Thanks for pointing out the typo. – Etienne Sep 26 '13 at 20:43