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 $2^{\mathbb N}$?
 A: 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.
A: 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
