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I know that there is something wrong with this proof (idea), however I could not find any mistake.

If I pick the open intervals $]q_n-\frac{\epsilon}{2^n}, q_n+\frac{\epsilon}{2^n}[$ (such that each $q_n$ is rational) as a covering for $\mathbb{Q}\cap[0,1]$ , then $m^*(\mathbb{Q}\cap[0,1])\leq\epsilon$. However since $\mathbb{Q}$ is dense , $\cup_{n\in \omega}{]q_n-\frac{\epsilon}{2^n}, q_n+\frac{\epsilon}{2^n}[}$ covers $[0,1]$, then $m^*([0,1])\leq\epsilon$. Furthermore, since $[0,1]$ is compact, it exists a finite subcover $\{U_i\}$, then $J^*([0,1])\leq \epsilon$ ($J^*$ is the outer Jordan ).

Thanks in advance.

user40276
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3 Answers3

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(...) since $\mathbb{Q}$ is dense, $\bigcup\limits_{n\in \omega}{]q_n-\frac{\epsilon}{2^n}, q_n+\frac{\epsilon}{2^n}[}$ covers $[0,1]$ (...)

Why is that? A simple proof that the set $\bigcup\limits_{n\in \omega}{]q_n-\frac{\epsilon}{2^n}, q_n+\frac{\epsilon}{2^n}[}$ DOES NOT cover $[0,1]$ is that its Lebesgue measure is at most $\epsilon$, hence for $\epsilon\lt1$, it cannot cover $[0,1]$.

Did
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Your claim that "since $\mathbb{Q}$ is dense, $$\cup_{n \in \mathbb{N}} (q_n - \epsilon/2^n, q_n + \epsilon/2^n)$$ covers $[0,1]$" is incorrect.

It is possible to construct a number not in $\cup_{n \in \mathbb{N}} (q_n - \epsilon/2^n, q_n + \epsilon/2^n)$. Look at the post

Constructing a number not in $\bigcup\limits_{k=1}^{\infty} (q_k-\frac{\epsilon}{2^k},q_k+\frac{\epsilon}{2^k})$

Community
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user Marvis mentioned that you can construct numbers that are not in the union of balls around each rational. This can be seen from a more abstract notion of the following: Let $A$ be dense in $X$, with $A$ countable. Then by enumerating each element of $A$ (call them $r_i$), is it true that $X\subset\cup_i B_{\epsilon_i}r_i$ for any decreasing sequence of $\epsilon_i$ with $\epsilon_i\rightarrow 0$? Suppose I consider all possible sequences of $B_{\epsilon_{i}}$ (more precisely, the unions $\cup_i B_{\epsilon_i}r_i$) . Then taking the intersection of all of these unions, I get back $A$ (since I can shrink balls around each $r_i$ to zero). Thus there must be a sequence of $B_{\epsilon_i}$ that covers $A$ but does not cover $X$.

Alex R.
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