# Some Questions about zero-dimensional subsets of the unit interval related to cantor set

Let $\mathbb{P}$ denote the set of all irrational numbers in the open segment$(0 , 1)$. let $K$ be the intersection of $\mathbb{P}$ and the standard cantor set and $H=\mathbb{P}-K$. as you know these sets are zero dimentional.I have three questions about these sets.

Q1.Is it true that the sets $K$ and $H$ are topologically Homeomorphic?

Q2.Is the space $K$ order isomorphic to $\mathbb{P}$ ?(I mean the existence of a monotonically increasing function from $K$ onto $\mathbb{P}$)

Q3.Is $H$ the union of countably many disjoint intervals from $\mathbb{P}$ ?

The standard "devil's staircase" Cantor function $f$ is a monotonically increasing function from $K$ onto $\mathbb P$. One way to see this is that a member $x$ of the Cantor set has a base-3 expansion $.d_1 d_2 \ldots$ where all $d_j \in \{0,2\}$, and $f(x)$ has the base-2 expansion $.b_1 b_2 \ldots$ where $b_j = d_j/2$. $x$ is irrational iff $d_1, d_2, \ldots$ is not eventually periodic iff $b_1, b_2, \ldots$ is not eventually periodic iff $f(x)$ is irrational.

• Ouch!! my answer came 20 mins late.. – Sungjin Kim May 9 '12 at 18:10
• Hello dear Robert. thank you very much for your description.This function is very beautiful and well-known. But How about The $Q1$ and $Q3$. what do you think about these Questions. Is it true that there is a homeomorphism between $H$ and $K$? – Ali Reza May 9 '12 at 18:10
• Thank you Dear !707107. How about you. Do you think that the answer of Robert Israel is complete? – Ali Reza May 9 '12 at 18:14
• I answered for Q3, not sure about Q1. – Sungjin Kim May 9 '12 at 18:18

For Q2, consider a mapping $\phi:K\rightarrow \mathbb{P}$ defined by $$\phi( \sum_{n=0}^{\infty} \frac{2a_n}{3^n})=\sum_{n=0}^{\infty} \frac{a_n}{2^n}$$ where $a_n$ is a sequence entirely consisted of 0 and 1, and not periodic.

For Q3, the answer is yes. H is just intersection of $\mathbb{P}$ and the complement of cantor set. The complement of cantor set is union of countably many disjoint open intervals.

• Thank you dear unKnown friend. You have also given the same beautiful and well-known answer. But the important part of this question is $Q1$.how do you think about it? – Ali Reza May 9 '12 at 18:24
• It will be a pain to write down, but Q1 can be solved by cutting P into countable pieces. Since both P and H are countable disjoint union of intervals having rational endpoints. They have to be homeomorphic. – Sungjin Kim May 11 '12 at 1:22

To answer Q1: In fact, there is an order preserving homeomorphism between K and H. Let [a_0,b_0]=[0,1], let (r_n) enumerate the rationals in [0,1], and for each finite sequence of 0s, 1s, and 2s starting with a 0 inductively define a closed interval so that the following properties are satisfied: a. Any interval contains uncountably many elements of K and has rational endpoints, b. Any two intervals corresponding to sequences of the same length intersect in at most endpoints and the one with the smaller (lexicographically) sequence is to the left c. In producing the intervals corresponding to sequences of length n+1, each interval has length at most 1/2^n and does not contain r_n as an interior point.

Then an infinite sequence of 0s,1s, and 2s starting with a 0 uniquely defines a point of K iff it’s not eventually all 0 or all 2 -- Just intersect the corresponding compact intervals of the finite subsequences, which always give a single point. The point will be a rational endpoint if eventually 0 or 2. Otherwise the intersection point isn’t ever an endpoint so can’t be rational since we avoided all rationals except for endpoints. But K is closed in the irrationals, in that if a sequence of elements of K converge to an irrational then that limiting irrational must be in K.

Finally, the same construction works for H and the desired homeomorphism just matches points with the same corresponding sequence. Continuity follows since small interval interiors go to small interval interiors.