# Determine if this set is compact or not

1. Jul 24, 2009

### Johnson04

1. The problem statement, all variables and given/known data
Let $$E$$ be the set of all $$x\in [0,1]$$ whose decimal expansion contains only the digits 4 and 7. Is $$E$$ compact? Is $$E$$ perfect?

2. Relevant equations

3. The attempt at a solution
My answer is: $$E$$ is compact and perfect.

By Heine-Borel theorem ($$E$$ is compact equivalent to $$E$$ is closed and bounded), since $$E$$ is bounded, and $$\forall x\in E$$, and $$\epsilon > 0$$, we always have that the neighborhood of $$x$$, $$N_\epsilon (x)$$ contains infinitely many elements of $$E$$, in other words, any number in $$E$$ is a limit point of $$E$$. Conversely, I assume $$E$$ is not closed, that is, there exists at least one number which is not in $$E$$, but it is a limit point of $$E$$. I denote this number by $$y$$. Suppose the i-th digit of $$y$$, say $$y_i$$ is the first digit which is not in $$\{4,7\}$$. Then we always can find some other numbers which are between $$y$$ and the closest number in $$E$$. So we can conclude that if $$y\not\in E$$, $$y$$ must not be a limit point of $$E$$. So we can conclude that $$E$$ is closed and perfect. By the aforementioned Heine-Borel theorem, it is also compact.

However, I am still worrying about this proof, since I am not very sure on the correctness of the proof. Can you guys help me to see if both my answer and my proof are correct? Thanks a lot!

2. Jul 24, 2009

### Dick

Your answer is correct. It's a Cantor type set. Your proof leaves a lot to be desired. And it is a bit complicated since you have to prove two things, that E is closed and perfect. Start with closed. As you said, if a point x is not in E, then it has a digit yi that is not 4 or 7. You have to show x is a positive distance away from E. Can you give an estimate for a positive lower bound on how far away it might be? Suppose the digit is say '3'?

Last edited: Jul 24, 2009