# Rudin's PMA Exercise 2.18 - Perfect Sets [duplicate]

I've been working through Chapter 2 questions and have thought about Exercise 2.18 for a while, but couldn't come up with an answer.

Is there a nonempty perfect set in R which contains no rational number?

I had a look here but I think this is wrong because obviously the Cantor set is not a subset of the rationals * (See below). Another thing that the above "solution" states is that the Cantor set only has endpoints but this is not true, e.g. 1/4 is a member of the Cantor set but is not an endpoint (See here).

Other solutions to this question based on a quick google search also give the same - seemingly incorrect - answer.

I have thought about sets like {x: Only 4 and 7 are in the decimal expansion of x, and the decimal expansion of x is non-repeating}, but this set doesn't work because 4/9 is a limit point of this set but is not a member of this set and so it is not a perfect set.

Other ideas I have come up with include starting with the irrational numbers in [0,1] and taking away lots of irrational numbers systematically, but I couldn't make this work.

The denseness of Q has something to do with it but I can't figure it out...

Any ideas?

*I say "obviously" but I should really explain myself here. The Cantor set is a perfect set, and every perfect set is uncountable, but any subset of rationals is countable.

## marked as duplicate by Adam Rubinson, Daniel Fischer, Amitesh Datta, Micah, Dan RustAug 24 '13 at 14:24

For any finite sequence of ones and zeros $f: n \to \{0, 1\}$, we'll define compact sets $F_f$ such that $F_{f \upharpoonright k} ⊇ F_f$ for any $k$. We'll also define $F_f := \bigcap_{n < ω} F_{f \upharpoonright n}$ for $f: ω \to \{0, 1\}$ and we want $diam(F_{f \upharpoonright n}) \to 0$ when $n \to ∞$ so $F_f$ is a singleton. Such construction gives obvious isomorphism between $\bigcap_{n < ω} \bigcup\{F_f: |f| = n\}$ and the Cantor space $2^ω$. This construction can be done in any $\mathbb{R}^n$.
In our case, $F_∅ := [a, b]$ for $a, b$ irrational. $U_f$ will be open subinterval of $F_f$ with irrational endpoints so $F_f \setminus U_f$ are two disjoint closed intervals. These will be $F_{f ∪ ⟨|f|, 0⟩}$ and $F_{f ∪ ⟨|f|, 1⟩}$. Since $F_∅$ and $U_f$ have irrational endpoints, by induction, all $F_f$ have irrational endpoints. So when $\mathbb{Q} = ⟨a_n: n ∈ ω⟩$ and $a_n ∈ F_f$ and $|f| = n$, we can choose $U_f$ so it contains $a_n$. And hence the constructed copy of Cantor set contains no rationals.