So the question is this:
Can we find a nice compact subset of the irrationals in $[0,1]$ that has close to full Lebesgue measure and then write down a simple formula or description for an irrational member of that compact subset?
Some background on where the question comes from:
Regarding Lebesgue measure, we clearly have $\mu([0,1]\setminus\mathbb Q)=1$. Let $\epsilon>0$ be small and $\{B_k\}_{k\in\mathbb N}$ be an open covering of the rationals in $[0,1]$ such that $\sum_{k=1}^\infty \mu(B_k)<\frac\epsilon2$. Thus since $\cup_{k=1}^\infty B_k$ is an open set and $[0,1]$ and $(\cup_{k=1}^\infty B_k)^c$ are both closed and bounded sets we get that $[0,1]\cap (\cup_{k=1}^\infty B_k)^c$ is compact.
Now we have that $$[0,1]\cap \left(\cup_{k=1}^\infty B_k\right)^c \ \subset \ [0,1]\setminus\mathbb Q \ \subset \ \left(-\frac\epsilon4,1+\frac\epsilon4\right)$$
And hence the set $[0,1]\setminus\mathbb Q$ contains a compact set of measure greater than $1-\frac\epsilon2$ and is contained in an open set of measure $1+\frac\epsilon2$. Thus we have a situation where $K\subset E\subset G$, $K$ is compact, $G$ is open, and $\mu(G)-\mu(K)<\epsilon$ so we know $E$ is Lebesgue measurable and its measure is arbitrarily close to 1, hence $\mu([0,1]\setminus\mathbb Q)=1$.
My question is about what the set $[0,1]\cap \left(\cup_{k=1}^\infty B_k\right)^c$ looks like. It only contains irrationals, but can we find a specific irrational in there?
What I want is a specific irrational number in that set, and even better than that would be a simple method for constructing such irrationals. I get that the rationals are removed in a careful way, and that it also removes uncountably many irrationals (of course leaving uncountably many irrationals too). But I have trouble figuring out precisely which irrationals are left behind.
Related, but this is not what I'm asking:
Now I know how to pick an irrational and then form a compact subset of the irrationals in $[0,1]$ that this particular irrational is a member of. Let $x$ be an irrational in $(0,1)$, and let $\{q_1,q_2,\ldots\}$ be the rationals in $[0,1]$. Then choose any $r_k\leq |q_k-x|$ for all $k$. We thus have that $x\in[0,1]\setminus \cup_{k=1}^\infty (q_k-r_k,q_k+r_k)$. I'd like to do this in reverse, pick the covering of $\mathbb Q$ and then figure out which irrationals are left behind.
My attempt at a construction of such an irrational:
Maybe using a specific value for $\epsilon$, e.g. $\epsilon=\frac12$, could help intuition here? Or maybe a specific enumeration of the rationals helps?
Just consider any arbitrary enumeration of the rationals in $[0,1]$ given by $\{q_1,q_2,\ldots\}$. And associate each one with a small positive rational radius $r_k$. Furthermore, make these radii small enough so that this construction works (we can make them all arbitrarily small since $\mathbb Q$ is of measure zero). For $I$ an arbitrary finite subset of $\mathbb N$ we know that $q_1+\sum_{k\in I} r_k$ is rational and thus appears somewhere in the list (again assuming the $r_k$ are chosen small enough). We also need that that $q_1\neq1$, or we can just pick any other rational besides $1$ to start the sum at (with the $r_k$ being sufficiently small). So now we can build a sublist of rationals $\{q_{k_1},q_{k_2},\ldots\}$ where $q_{k_1}=q_1+r_1$. Then we add the radius associated with $q_{k_1}$ to this to get $q_1+r_1+r_{k_1}$. This is again rational so we have $q_{k_2}=q_1+r_1+r_{k_1}$ and pick the radius associated with this to get $q_{k_3}=q_1+r_1+r_{k_1}+r_{k_2}$, etc.
Thereby we generate an infinite series $x=q_1+r_1+\sum_{n=1}^\infty r_{k_n}$ which by necessity converges (increasing and bounded from above) and cannot be rational. If $x$ were rational, then we would have a radius associated with it and would just add it to the sum, but we have already done this ad infinitum. As long as the initial (rational) radii $r_k$ are chosen sufficiently small (we can always choose them arbitrarily small by the $r_k<\epsilon/2^k$ trick with $\epsilon>0$ arbitrary and sufficiently small), then this works and gives us an irrational is in $(0,1)$.
Can all irrational numbers that remain uncovered by the covering a rationals be constructed in this way though? I think it should at least be countable infinite amount of irrationals that can be constructed by this infinite series of radii technique, but if the full uncountable amount of irrationals that remain can be approximated this way I do not know. After thinking about it some more I suspect the answer is: "No, even with all possible such series constructions there will always be uncountable infinity of irrationals left undiscovered."
Let $\epsilon=\frac12$ and choose the rational enumeration $\left\{0,1,\frac12,\frac13,\frac23,\frac14,\frac34,\ldots\right\}$ just going along by enumerating the denominators and discarding any repeats. And let $B_k=\left(q_k-\frac\epsilon{2^k},q_k+\frac\epsilon{2^k}\right)$ for each $q_k$ in this list of rationals in $[0,1]$. Then we have $\mu([0,1]\cap \left(\cup_{k=1}^\infty B_k\right)^c)=\frac38$ and it only contains irrationals and is compact. Can someone write down a formula for an irrational in this set? I get that $x=0+\frac14+\frac1{128}+\cdots$ by using the above method. But I haven't figured out how to come up with a formula for the general term of this series. It is probably possible to describe it using a sequence of mutually prime numbers for each particular denominator, but this might just give an ugly formula so I didn't pursue it further. Is there a nicer way? Can we get a nice formula?
Is there a very nice representation of such an irrational $x$ constructed in the above manner or using another probably more efficient construction?
A couple of related MathSE threads: