I'm having some weird misunderstanding here that I can't resolve in my head. Let $\{q_1,q_2,\ldots\}$ be an enumeration of the rationals. Then $\cup_{i=1}^\infty(q_i-\epsilon2^{-i-1}, q_i+\epsilon2^{-i-1}) \supset \mathbb{R}$ since the rationals are dense. Yet $$\lambda(\cup_{i=1}^\infty(q_i-\epsilon2^{-i-1}, q_i+\epsilon2^{-i-1})) < \sum_{i=1}^\infty \lambda(q_i-\epsilon2^{-i-1}, q_i+\epsilon2^{-i-1}) = \epsilon$$ where $\lambda$ is the Lebesgue measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ Yet we know $\lambda(\mathbb{R}) = \infty$, so by monotonicity of the measure we should have $$\lambda(\cup_{i=1}^\infty(q_i-\epsilon2^{-i-1}, q_i+\epsilon2^{-i-1})) \geq \lambda(\mathbb{R}) = \infty$$ which contradicts what we had before. I'm definitely missing something here, but can't figure out what it is. Any clarification would be greatly appreciated, thanks in advance!
-
Should ${q_1,\cdots}$ be an enumeration of the rationals? – KReiser Sep 24 '13 at 05:06
-
The reals are uncountable, hence cannot be enumerated. – Calvin Lin Sep 24 '13 at 05:06
-
$\mathbb{Q}$ is dense in the reals, but that doesn't imply that any open set containing $\mathbb{Q}$ must equal the reals. Any closed set containing them must, of course. – copper.hat Sep 24 '13 at 05:07
-
Why do you believe $\mathbb{R}$ should be a subset of the union you propose? For any $r\in\mathbb{R}/\mathbb{Q}$ I can give you an iteration $q_i$ of the rationals such that $r$ isn't in the union-of-intervals you construct... – Steven Stadnicki Sep 24 '13 at 05:08
-
Modulo a few facts of measure theory, you have just given a proof that the reals cannot be enumerated. – André Nicolas Sep 24 '13 at 05:09
-
What you have shown above is that any countable set has Lebesgue measure zero. – copper.hat Sep 24 '13 at 05:10
-
Sorry I fixed the typo, the enumeration is of the rationals. Sorry about that! – anegligibleperson Sep 24 '13 at 08:33
2 Answers
It's not necessarily the case that $\mathbb{R}\subset S_\epsilon=\cup_{i=1}^\infty(q_i-\epsilon2^{-i-1}, q_i+\epsilon2^{-i-1})$. In fact, for any fixed real $r\in\mathbb{R}\backslash\mathbb{Q}$ it's easy to construct a specific enumeration of the rationals such that $r\not\in S_\epsilon$: for each rational $q$, enumerate it 'late enough'; that is, let $q$ be $q_i$ for $i$ large enough that $|r-q|\gt 2^{-i-1}$. There are plenty of rationals 'far enough away' that we can use them to fill in any gaps the enumeration between the rationals that are closer to $r$.
In fact, for any finite set $A\subset\mathbb{R}\backslash\mathbb{Q}$ of non-rational reals, we can build an enumeration of the rationals that misses all the reals in $A$. (Easy exercise: show that 'finite' is the best we can do; that is, there's some countable set $A$ of non-rationals such that it's impossible to miss all of the reals in $A$)
- 51,746
If $A$ is dense in $\mathbb R$, then $\mathbb R$ is the closure of $A$, hence $\mathbb R$ is the only closed set containing $A$. However, this does not imply that every open set containing $A$ is $\mathbb R$. Your measure theoretic observation proves that the open sets you describe do not cover $\mathbb R$. Some simpler examples of open sets between $\mathbb Q$ and $\mathbb R$ are $\mathbb R\setminus\{\sqrt 2\}$, or $\mathbb R\setminus(\text{any finite set of irrational numbers})$, or $\mathbb R\setminus\{\sqrt 2+k:k\in\mathbb Z\}$. In general, removing any closed set of irrational numbers leaves you with an open cover of the rational numbers.
- 53,602