# Do almost all permutations of the interval have dense graphs?

I'm thinking about bijections from $I=[0,1]$ to itself. It's clear that, cardinality-wise, almost all of them are, for example, discontinuous. I'm wondering how badly discontinuous. My intuition tells me that most bijections on the interval have graphs that are dense in the square $I\times I$, but I'm not sure how to prove it.

If we choose an arbitrary rectangle $[a,b]\times[c,d]\subset I\times I$, it should suffice to show that some $x\in [a,b]$ gets mapped into $[c,d]$. My natural thought is some kind of probability argument, but since these points come from a permutation, they're not actually random, so I don't think I get to make such an argument. Nevertheless, out of all permutations, "most" of them should avoid gaps, I think.

How can I formalize this idea?

Edit: I should clarify my question. I know there are multiple ways to talk about "most" of an infinite set. I'd like to hear anything intelligent about permutations whose graphs are dense versus those that aren't. I'd take: This kind counts as 'most', in terms of: cardinality, measure, Hausdorff dimension, category, or any other commonly used way of talking about "most" of a set. No, I haven't got a specific topology in mind. If you do, I'd love to hear about it. If permutations whose graphs are dense sets are actually kind of "rare", in some sense, I'm curious about that, too.

Sorry if this makes my question too general, but I think it's specific enough to allow for helpful answers.

• The answer depends on what you mean by "most bijections". Do you have some sort of measure or topology in mind? Or do you mean cardinality-wise? – Servaes May 14 '18 at 9:18
• Cardinality-wise doesn't work: obviously there are already $\mathfrak{c}^{\mathfrak{c}}$ mappings from $[0, 1/2]$ into itself, and you can extend all of these with the identity $(1/2, 1]$ into itself. – Mees de Vries May 14 '18 at 9:25
• @Servaes, see my edit – G Tony Jacobs May 14 '18 at 9:36
• @GTonyJacobs There is no widely accepted notion of measure, topology, dimension or any other way of interpreting "most functions" on the set of all bijections $X\to X$ except for the cardinality which in that case doesn't really give you much. – freakish May 14 '18 at 10:36
• I was aware of it. I haven't mean to indicate that I'm not.... or that I wasn't. I haven't been dishonest in any way, but I'll plead guilty to scatter-brained, and now disoriented. When I asked, "Is it at least clear that both kinds have the same cardinality?," I wasn't talking about continuous versus discontinuous. I was asking about dense graphs versus not dense graphs. When I phrased my last comment as a question, it was a common rhetorical way of expressing - "I assume we're on the same page about this fact." I'm not trying to be disagreeable. I respect you very much, @mercio. – G Tony Jacobs May 14 '18 at 11:02