Is there a bounded, closed, uncountable subset of irrational numbers

Question

We can easily find examples of closed and uncountable subset of irrational numbers. What about if we add the condition of the set being bounded?

My first initial reaction was that such a set of not possible. Various approaches to proving it were using the fact that since it's closed and bounded, it must be compact and then claiming something using the existence of a finite subcover. But this hasn't led me to anything.

Is there an example of such a set? If not can we prove why there can't be such a set?

Answer 1

Here is an alternative to the Cantor set examples, but it requires some measure theory. We can construct an open set of arbitrarily small measure that contains all the rationals by enumerating the rationals and taking the union of open intervals of length $2^{-n}\epsilon$ around the $n$th rational. The complement of this set is then a closed set containing only irrational numbers. If we want a bounded set, just intersect if with a bounded closed interval. As long as that interval's length is larger than $\epsilon$, we're guaranteed to end up with a set of positive measure (and such a set must be uncountable).

On a related note, we can show that if we have an unbounded, uncountable closed set of irrationals then we can construct a bounded set from it. Just take its intersection with each of the unit intervals $[n,n+1]$ for $n\in \mathbb Z$. Since there are only countably many such intervals, at least one of those intersections has to be uncountable.

Answer 2

I have constructed an example

We consider the set $X=\{0.a3bc3def3,\dots:~\text{where} ~a,b,c,d,.... =1~\text{or}~2\}$

Here

 the set is bounded.

 the set contains only irrational numbers.

 bounded as each number is $<1$

 the set is uncountable

 and it is obvious that this is closed.