Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

let $Q$ := $\{q_1, ...,q_n,...\}$ be an enumeration of the rationals.

let $\epsilon_n$ = $\frac{1}{2^n}$ and let $S_i$ = $(q_i-\epsilon_i, q_i + \epsilon_i)$

Let $S$ be the union of all $S_i$. Take the complement of $S$ in $\mathbb{R}$.

I have a few queries about this set. I'm sure they are all somewhat trivial but i've thought a bit about them and the penny still hasn't dropped.

Firstly, by considering a geometric series $S$ has maximum length 2. So the complement is infinte. But the complement has no intervals, it's just irrational points. And i can't think of an example of one of these points. If we have an irrational in the complement it is "between" two sets, with rational endpoints. these rational endpoints should then have another open set about them, which even if infinitesimal, should cover "one" point, our irrational element of the complement. So now our one example of the complement is no longer in the complement.

Secondly, once armed with the intuition about what this set really looks and feels like, i wonder about the fact it has infinite lenght. If the set of irrational elements of the set are wedged in between a countable number of open sets, it feels like there should be a countable number of points in the complement if $S$. But then the length would be zero.

To phrase the second part of the question more appropriately maybe i could then redefine $\epsilon_n$ as $\frac{\pi}{2^n}$, and try and argue since the $S_i$ now have irrational endpoints, the only irrationals in the complement can be the endpoints, and this shows they are countable...

where am i going wrong?

share|cite|improve this question
You don’t have a geometric series of lengths; you need to replace your $\epsilon_n=\frac1n$ by $\epsilon_n=2^{-n}$ in order to get what you wanted. – Brian M. Scott Mar 5 '13 at 0:05
Did you mean $\epsilon_n=\frac1{2^n}$ rather than $\frac1n$? – Asaf Karagila Mar 5 '13 at 0:05
silly. of course that's what i meant. i'll switch them now. thanks. – pad Mar 5 '13 at 0:09
up vote 1 down vote accepted

First of all, note that you don't specify the intervals. You assert their existence, but you don't give an exact definition. This is also true for the enumeration of the rationals.

If $q_0=0$ the irrationals which do not appear in this set are different from those where $q_0=500^{100000^{100}}$. So specifying an irrational number which is not in $S$ is equally impossible.

Lastly, the complement of a set of measure $2$ need not contain intervals. Arguing against this is easily reduced to an argument against the following mathematical truth:

Let $\Bbb I$ denote the irrational numbers, then $\Bbb I$ is the intersection of countably many open sets, but it contains no proper interval.

This is because the irrational numbers form a totally disconnected space, the connected sets are singletons (do not confuse this with the space being discrete! The singletons are not open!) and so it is perfectly fine.

Now seeing how you covered all the rational numbers, the complement must lie within the irrational numbers, and since those are totally disconnected it cannot be anything else itself.

To your last argument, it is not true that every irrational is the endpoint of some interval, indeed there are many of them which are not the endpoints of any interval. They are accumulation points of the endpoints of the intervals, though. And a countable set can easily have an uncountable set of limit points (e.g. the rational numbers are dense in $\Bbb R$).

Also related: Formalizing an idea, and if one is bored enough then one can read my back-and-forth comments with a well-known crank.

share|cite|improve this answer
thank you. this answer is brilliant. is there an intuitive way of seeing the flaw with saying; since the irrational points in the complement are surrounded by open sets in $S$ there is "at most" as many irrationals in the complement as there are open sets in $S$...therefore the irrationals in the complement are countable? or should one simply accept this as foundation? – pad Mar 5 '13 at 0:43
@pad: Yes, the intervals do not cover all the real numbers, only the rational numbers. And while being dense in terms of order, they are not very dense in terms of volume. – Asaf Karagila Mar 5 '13 at 1:21
@Asaf: That back and forth was...frustratingly entertaining. – Cameron Buie Mar 5 '13 at 1:30
@Cameron: You can imagine how it was to participate in that argument! :-) I liked its speedy conclusion, though. If one cannot stand the idea of accepting the axioms of ZFC, one should not be trying to formulate things in them. – Asaf Karagila Mar 5 '13 at 1:34
it was hardly an argument. thanks again. – pad Mar 5 '13 at 9:19

Here's a different example that might help 'picturing' things. Consider instead the set of open balls $S_i = (q_i - |q_i - \pi|, q_i + |q_i - \pi|)$. This is an open covering of the rationals whose complement is $\{ \pi \}$.

This was the easiest way to see it, but we could still construct a covering using your recipe that doesn't contain $\pi$, either by carefully choosing the enumeration, or by a small variation: picking $\epsilon_i = \min\{ |q_i - \pi|, 2^{-i} \}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.