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I would like a formal statement for the following fact.

Let $\gamma$ be an analytic curve from the open interval $(-1, 1)$ to $\mathbb{R}^2$. With probability $1$ (in some sense), the image of the curve does not intersect $\mathbb{Q}^2$.

How can I do it?

frv
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    The support of a function is a subset of its domain. You probably meant the image. – jjagmath May 28 '22 at 17:39
  • I'm not a measure theory person, but have you thought of the simpler problem of $f:(-1,1)\to\mathbb R$, and $f((-1,1))$ not containing any rational with probability 1? – David P May 28 '22 at 17:40
  • Probability is the wrong way to approach this. Do you know about Baire category? – Moishe Kohan May 28 '22 at 17:41
  • @jjagmath yep that's what I meant, thanks :) – frv May 28 '22 at 17:41
  • @DavidP yep that does not work, if you join two irrationals by a curve you hit rationals. But in higher dimensions the situation looks very different – frv May 28 '22 at 17:43
  • @MoisheKohan I know Baire category theorem. Can you please elaborate? – frv May 28 '22 at 17:50
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    Can you please elaborate? --- It's enough to show (for some distance function making the set of analytic curves into a complete metric space) that, for each $p \in {\mathbb Q}^2,$ most curves (in the Baire category sense; i.e. a co-meager set of curves) do not pass through $p.$ (Indeed, probably much more can be shown, namely that a co-nowhere-dense set of curves has this property, although after the next step this additional strengthening vanishes.) This is because you can then take the countable intersection (continued) – Dave L. Renfro May 28 '22 at 18:20
  • of the various co-meager sets (one co-meager set for each $p \in {\mathbb Q}^2)$ to get a co-meager set of curves each of which does not pass through any point of ${\mathbb Q}^2.$ See the paragraph beginning with "An interesting feature of typical properties" in my answer to Generic Elements of a Set. – Dave L. Renfro May 28 '22 at 18:20
  • Incidentally, a similar "countable intersection" result can be used to show that "most translations" of the Cantor middle thirds set (or any perfect nowhere dense set, even one with positive measure) contain no rational numbers. For even stronger results and references, see this 11 May 2000 sci.math post. Thus, if you want a Cantor set consisting only of irrational numbers, all you need to do is "Baire-randomly translate" the Cantor middle thirds set (or any Cantor set what-so-ever)! – Dave L. Renfro May 28 '22 at 18:27
  • @DaveL.Renfro Baire category argument works for continuous curve. How to adjust that to analytic curves? Do they form a closed subset of continuous curves? Also how to deal with open interval? – 183orbco3 Jun 12 '22 at 04:20
  • @183orbco3: Your questions are why I said "for some distance function making the set of analytic curves into a complete metric space" (i.e. used "some", rather than saying something more definite and specific). A standard metric is the metric associated with uniform convergence on compact subsets, but I don't remember much about that metric to know whether it would be suitable. However, if this metric or some other reasonably well known metric doesn't work (continued) – Dave L. Renfro Jun 12 '22 at 07:39
  • then maybe some artificially constructed metric can be found to do the job, although this would likely be a nontrivial and fairly creative task. For example, the set of Besicovitch functions is known to be a meager subset of the space of continuous functions (sup metric), and thus the Baire category method for existence doesn't work for the sup metric, but Where the continuous functions without unilateral derivatives are typical by Malý (1984) introduces a different metric for which the method does work. – Dave L. Renfro Jun 12 '22 at 07:46

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