Do almost all permutations of the interval have dense graphs?

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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.