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:
- 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 $\tau$?)
- Topological (e.g. is there a natural topology on the power set of $\mathbb R$ where $\tau$ is meagre, or even nowhere dense?)
- 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?
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.