It seems intuitive to believe that most subsets of $\mathbb R$ are neither open nor closed.

For instance, if we consider the collection of all (open, closed, half-closed/open) intervals, then one can probably make precise the notion that "half" of all intervals in this collection are neither open nor closed. (Whether this will amount to a reasonable definition of what it means for most subsets to be neither open nor closed might be up for debate.)

If this intuition is correct, is there a way to formalise it? If not, how would we formalise its being wrong?

To be clear, I am happy for a fairly broad interpretation of the term "most". Natural interpretations include but are not limited to:

  1. Measure-theoretic (e.g. is there a natural measure on (a $\sigma$-algebra on) the power set of $\mathbb R$ that assigns negligible measure to $\tau$?)
  2. Topological (e.g. is there a natural topology on the power set of $\mathbb R$ where $\tau$ is meagre, or even nowhere dense?)
  3. Set-theoretic (e.g. does the power set of $\mathbb R$ have larger cardinality than $\tau$?)

Here, $\tau$ is (obviously) the Euclidean topology.

Actually, that last version of the question in parentheses might have the easiest answer: Let $\mathcal B$ be the Borel sets on $\mathbb R$. We have that $|\tau| \le | \mathcal B | = | \mathbb R | < \left| 2^{\mathbb R} \right|$. (For details on the equality, see here. For a much simpler proof, see this answer.)

Are there alternative ways to make this precise?

  • $\begingroup$ A trivial answer to 1. and 2. would be to define a topology on $P(\Bbb R),$ the power-set of $\Bbb R$, as $\{\emptyset, P(\Bbb R)\setminus \tau, P(\Bbb R)\}$ and a measure $m$ with $m(P(\Bbb R))=1$ and $m(\tau)=0.$ But I don't think this is quite what you're hoping for. $\endgroup$ – DanielWainfleet Sep 16 '18 at 1:12
  • $\begingroup$ @DanielWainfleet Indeed. I was hoping the word "natural" would be enough to rule out such answers. Unless there are other reasons I'm not seeing that would make such a definition natural? $\endgroup$ – Theoretical Economist Sep 17 '18 at 22:43

Since each open non-empty subset of $\mathbb R$ can be written has a countable union of open intervals and since the set of all open intervals has the same cardinal as $\mathbb R$, the set of all open subsets of $\mathbb R$ has the same cardinal as $\mathbb R$. And since there is a bijection between the open subsets of $\mathbb R$ and the closed ones, the set of all closed subsets of $\mathbb R$ also has the same cardinal as $\mathbb R$. So, in the set-theoretical sense, most subsets of $\mathbb R$ are neither closed nor open.

  • $\begingroup$ Thanks; I am aware of this fact (see the bottom of my post). However, I like your proof. It is much simpler than mine. $\endgroup$ – Theoretical Economist Sep 15 '18 at 17:43
  • $\begingroup$ @TheoreticalEconomist I wonder whether someon will provide a measure-theoretical or a topological answer. $\endgroup$ – José Carlos Santos Sep 15 '18 at 17:44
  • $\begingroup$ Yes, that would certainly be much more interesting. $\endgroup$ – Theoretical Economist Sep 15 '18 at 17:47
  • $\begingroup$ To the proposer: The set of bounded open real intervals with rational end-points is a countable base (basis) for $\tau.$ For any topology $T$ with a countable base $B$ we have $|T|=|\{\cup C: C\in P(B)\}|\leq |P(B)|\leq 2^{\aleph_0}=|\Bbb R|.$.... So $|\tau|\leq |\Bbb R|.$ And we also have $|\tau|\geq |\{(r,r+1):r\in \Bbb R\}|=|\Bbb R|.$ $\endgroup$ – DanielWainfleet Sep 16 '18 at 1:27

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