The question:

Let us enumerate the set of rationals $\mathbb Q = \{r_n\}_{n=1}^{\infty}$ and define \begin{equation} B = \bigcup_{n=1}^{\infty} \left(r_n - \frac 1 {2^n}, r_n + \frac 1 {2^n}\right)\,. \end{equation} Show that the outer measure $m^*(B) < \infty$ and therefore $B\subseteq \mathbb R$.

If $A = B^c$, show also that $A$ is closed, $A$ does not contain any empty open interval and that $m^*(A) = \infty$.

An attempt at an answer:

Let $B_n = \left(r_n - \frac 1 {2^n}, r_n + \frac 1 {2^n}\right)$. Then according to the definition of length $\ell$, as $B_n$ is an open interval, \begin{equation} \ell(B) = r_n + \frac 1 {2^n} - \left( r_n - \frac 1 {2^n} \right) = \frac 2 {2^n} = \frac 1 {2^{n-1}}\,. \end{equation} Therefore \begin{equation} m^*(B) = \inf\left\{\sum_{n=1}^{\infty} \ell(B_n)\right\} = \inf\left\{\sum_{n=1}^{\infty} \frac 1 {2^{n-1}}\right\} = 2, \end{equation} since the series in question is a geometric series that converges to $2$. Therefore $m^*(B) < \infty$ and according to definition, $B$ has to be of the form $(a,b)$, where $a,b\in\mathbb R$. Therefore $B\subseteq \mathbb R$.

As for the set $A = B^c$, since $B$ was open according to the above, $A$ is closed by definition. However this is as far as I got. I tried assuming that $A$ does contain an open unempty interval, but that got me nowhere:

If $A$ did indeed contain such an interval $I = (c,d) \subset A$, there would be an $a\in A$ and a $\delta > 0$, with which $(a - \delta,a + \delta)\subseteq I$. Also, in this case $0 < \ell(c,d) = d - c < \infty$. Maybe my understanding of the situation is lacking, but to me this doesn't seem to imply anything meaningful and I can't seem to think of a direct proof.

Any help would be much appreciated.


Your proof for the outer measure being finite is good. There are some errors in what you write after, though.

$B$ is certainly not of the form $(a,b)$. This is easily shown, assume that it is of the form $(a,b)$ for some $a,b\in\mathbb{R}$. Then we can take the ceiling of $b$, $\lceil b\rceil$, which must be rational and is greater than $b$. But all rational points are in $B$, a contradiction.

The reason $B$ is open is that it is the union of open sets, which is open by definition of a topology. To see that $A$ cannot contain a nonempty open subset, assume it does. Then, $\forall a\in A$ we have that there is a neighborhood of $a$ contained entirely in $A$, we denote this neighborhood $(a-\epsilon,a+\epsilon)$ for some $\epsilon>0$. By the density of the rational numbers in $\mathbb{R}$, $\exists q\in \mathbb{Q}$ such that $q\in(a-\epsilon,a+\epsilon)$. But, since $q$ is rational, we have that $q\in B$ which is the complement of $A$, a contradiciton. Thus such a nonempty open subset of $A$ cannot exist.

  • $\begingroup$ I'm not sure what a topology is, but we did prove that an enumerable union of open sets must be open, using the definition of an open set. $\endgroup$ – SeSodesa Mar 12 '18 at 14:02
  • $\begingroup$ How can you talk about open sets without a topology (rhetorical question, of course this isn't up to you)? That is a weird way to run the class $\endgroup$ – Tony Mar 12 '18 at 14:20
  • $\begingroup$ Well, in our intro to real analysis we simply defined an open set of real numbers as a set that only has interior points, without ever using the word ""topology", even if it had been appropriate. $\endgroup$ – SeSodesa Mar 12 '18 at 14:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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