Perfect set of irrationals

Question

I solved the following exercise:

Let $\{r_1,r_2,r_3,\dots \}$ be an enumeration of the rational numbers and for each $n \in \mathbb N$ let $\varepsilon_n = 1/2^n$. Define $O = \bigcup_{n=1}^\infty V_{\varepsilon_n}(r_n)$ and let $F=O^c$.

(a) Argue that $F$ is closed and nonempty consisting only of irrational numbers.

(b) Does $F$ contain any nonempty open intervals? Is $F$ totally disconnected?

However I am now stuck with (c):

(c) Is it possible to know whether $F$ is perfect? If not, can we modify this construction to produce a nonempty perfect set of irrational numbers?

I tried and I can't do it but I would really like to know the answer. Of course it is not possible to know that $F$ is perfect because it might contain isolated points but can there be a perfect set of irrational numbers?

Do you means all number in the set must be irrational, or you means the set must contains all irrational number? — Gina, Jan 05 '14 at 13:32
Gina: he means the second one. Gabriel: $V_{\epsilon_n}(r_n)$ and $O^c$ are baby-Rudin notation for "the ball of radius $\epsilon_n$ about $r_n$" and "the set-compliment of $O$" respectively. — Ben Grossmann, Jan 05 '14 at 13:41
@Gina I mean that it should be a possibly strict subset of the irrational numbers. — newb, Jan 05 '14 at 14:56
@GabrielR. It denotes the open ball around $r_n$ or radius $\varepsilon_n$ and the other is the complement of $O$ in $\mathbb R$. — newb, Jan 05 '14 at 14:57
https://math.stackexchange.com/questions/1975132/is-this-a-perfect-set?rq=1 — Alain, Jun 17 '17 at 11:51
@Gina : I believe the set of all Irrationals is not a perfect set. — MSIS, Nov 15 '22 at 22:13

Ben Grossmann · Answer 1 · 2014-01-05T14:18:52.503

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Note that the existence of an isolated point in this construction requires a bit of a remarkable coincidence where two neighborhoods happen to share a border.

Note however, that by our construction, that border must be rational, which means that it is contained by some other neighborhood in $O$.

edited Jan 05 '14 at 14:18

answered Jan 05 '14 at 13:46

Ben Grossmann

225,327

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The boundary points of all the $V_{\varepsilon_n}(r_n)$ are rational, so if two such sets have a common boundary point, that does not belong to $F$. The problematic situation is when the the boundary points of the $V_{\varepsilon_n}(r_n)$ approach an irrational $x$ from both sides, can it happen that $O$ contains $(x-\delta,x)\cup (x,x+\delta)$ for some $\delta > 0$? – Daniel Fischer Jan 05 '14 at 14:15
@DanielFischer I'm not sure how I missed that. Thanks. – Ben Grossmann Jan 05 '14 at 14:19
@BenGrossman: Is being a perfect set a Topological property? – MSIS Nov 15 '22 at 22:14
@MSIS A set is "perfect" if it is closed and has no isolated points. – Ben Grossmann Nov 15 '22 at 22:24
@Ben Grossman: So it seems those properties are preserved by homeomorphism, which is a closed map. Similar for isolated points. – MSIS Nov 15 '22 at 22:46
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@MSIS I seem to have misinterpreted your question. The answer to your original question is indeed "yes". – Ben Grossmann Nov 15 '22 at 22:49
@BenGrossmann: Thanks. – MSIS Nov 15 '22 at 22:51

Perfect set of irrationals

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