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In some of his writings, Paul Cohen gave an informal, motivational discussion about the word generic (as it is used in forcing). While very suggestive, the discussion leaves the meaning of the word ambiguous, and could lead someone to guess that it is related to probability theory. Indeed, there was a recent MO question along these lines. As the comments and answers to that question make clear, generic is actually closely related to Baire category and not to measure theory. As one learns in an analysis course, it is perfectly possible for a comeager set to have measure zero and for a meager set to have positive measure.

This got me wondering. Is there a simple/natural example of an independence result where, if you were to appeal to probabilistic intuition, you would guess wrongly what happens, because the generic event has probability zero? A good example should (A) be an independence result that is natural-sounding and easy for a non-set theorist to understand and (B) have an obvious and tempting—but wrong—line of reasoning based on conflating measure with category.

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    $\begingroup$ Take a meager co-null set (e.g. the union of fat Cantor sets with measures approaching to $1$), then adding a Cohen real would avoid this set. Now take some "natural description of a Cohen real" and make that into your wanted example. $\endgroup$
    – Asaf Karagila ♦
    Commented Jan 18, 2021 at 23:12
  • $\begingroup$ Sorry to be dumb here, but what do you mean by an independence result? Are we talking about set-theoretic independence here? $\endgroup$
    – Anthony Quas
    Commented Jan 19, 2021 at 5:25
  • $\begingroup$ @AnthonyQuas : Yes, I mean set-theoretic independence. $\endgroup$
    – Timothy Chow
    Commented Jan 19, 2021 at 5:40
  • $\begingroup$ Hmmm so I'm not sure about the set-theoretic independence part, but in case it's relevant, the set $A=\{x\in[0,1]\colon \exists \text{infinitely many }p,q\text{ such that }|x-\frac pq|<\frac{1}{q^3}\}$ is an easily understandable residual set of measure 0. $\endgroup$
    – Anthony Quas
    Commented Jan 19, 2021 at 6:24
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    $\begingroup$ I think the way the question is framed is a bit misleading. There is not just one “meaning” of generic that is close to comeagerness. “Generic” has only a technical meaning in terms of a partial order and a collection of dense subsets of it. It can correspond to comeagerness, or Lebesgue measure one, or these notions relativized to some other topology or measure, or “measure one” for some other ideal like the Marczewski ideal, or generic with respect to ideals on higher Baire space, or a generic closed subset of a stationary set, etc. etc. $\endgroup$
    – Monroe Eskew
    Commented Jan 19, 2021 at 9:59

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If the Borel-Cantelli lemma counts as probabilistic intuition, then here's an example. Think of the real $x$ that you adjoin to a ground model as a sequence of $0$'s and $1$'s, and let $f(n)$ be the length of the $n$-th run of consecutive $1$'s in $x$. If the bits in $x$ were chosen by independent flips of a fair coin (or even of a biased coin as long as both sides of the coin have positive probability), then the inequality $f(n)>n$ would (with probability $1$) hold for only finitely many $n$ (by Borel-Cantelli). But for a Cohen-generic $x$, that inequality holds for infinitely many $n$. In fact, for any function $g:\omega\to\omega$ in the ground model, $f(n)>g(n)$ for infinitely many $n$.

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  • $\begingroup$ This looks promising. Can this be parlayed into a natural-sounding statement of the form, "such-and-such a statement is unprovable in ZFC"? $\endgroup$
    – Timothy Chow
    Commented Jan 19, 2021 at 2:58
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    $\begingroup$ @TimothyChow: That seems a bit unlikely, because (at least as I understand it) the claim “genericity is more connected to Baire category than measure” holds just for the specific forcing poset Cohen first used (and related forcing posets), not about forcing in general. You can also force to adjoin a random real, using the poset of Borel sets modulo measure zero. So the category/measure distinction can certainly be seen in Cohen forcing specifically, as this answer shows, but I don’t see how we’d expect it to show up in an independence result. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Jan 19, 2021 at 11:55
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Freiling's Axiom of Symmetry (discussed also here) posits that for any function $f$ from $[0,1]$ to the at most countable subsets of $[0,1]$ there exists a pair of points $x$ and $y$ such that $x\notin f(y)$ and $y\notin f(x)$. Freiling argued that, for a "random" $f$, $x$, and $y$, the probability that both $x\notin f(y)$ and $y\notin f(x)$ is 1. However, the axiom is actually independent of ZFC (and actually equivalent to $\neg$CH).

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    $\begingroup$ I knew about Freiling's axiom but I don't think it's quite what I had in mind. This isn't an example of mistakenly interpreting "generic" as "probability 1." $\endgroup$
    – Timothy Chow
    Commented Jan 19, 2021 at 0:34
  • $\begingroup$ @TimothyChow No, it's not quite the same, you're right. It does seem like a relevant point of comparison though, since it expressed a converse situation where what looks like a "probability 1" situation is not (necessarily) generic. $\endgroup$
    – Buzz
    Commented Jan 19, 2021 at 1:24

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