is $\{ x \in \mathbb{R} : 0\leq x \leq 1$ and $x$ is irrational $\} $ compact?

Question

is $\{ x \in \mathbb{R} : 0\leq x \leq 1$ and x is irrational $\} $ compact ?

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Theorem Let $A \subset \mathbb{R}^n$ the following are equivalent

• $A$ is closed and bounded
• Every open cover of $A$ has a finite subcover
• Every sequene in $A$ has a subsequence which converges to a point of $A$

A subset of $\mathbb{R}^n$ satisfying one (and hence all) of the conditions of Theorem is called compact

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(would take like more than hour to draw that on latex by the way)

I thought at first it was compact it was throwing me off that points are closed and thought all the irrational points togheter was a closed set which is not the case.

I know this because looked at back of book of marsden "elementary classical analysis" on ex 3.1 1.a. It just says it is not compact but dosent elaborate why.

For it to not be compact it needs to fail one of the conditions of the theorem. It is not obvious to me a sequence made up of irrationals within [0,1] that converges to a point outside the given set.

can make a loose argument how second bullet point of theorem fails.could make up an open cover consisting of union of open balls at the irrational points of the set with the radius of the smallest distance bw a point and its neighbor. and then say there is no finite subcover since the open cover is made up of infinite sets.

wondering what is the most elegent way to put it!

Answer 1

That set is not closed, notice that $0$ is a limit point for the set and is not in the set.

Answer 2

(i). $A=[0,1]$ \ $Q$ is not closed because if $ x\in A $ then the sequence $(x/n)_{n\in N}$ of points of $A$ converges to $0, $ and $0\not \in A.$ In fact the closure of $A$ is $[0,1].$

(ii). $C=\{(2^{-n},2^{1-n}) :n\in N\}$ is an infinite open cover of $A$, and no proper subset of $C$ is a cover of $A.$

(iii). $A$ is not closed. Let $T= (t_n)_{n\in N}$ be any sequence in $A$ converging to a point not in $S$. Then $T$ cannot have a subsequence converging to anything else, hence no subsequence of $T$ converges to a member of $A.$

Remark. The fact that $A$ is dense in $[0,1]$ follows at once from the def'n of the real number system $R$. Without a def'n of $R$ you can't prove anything about it.