Closed, uncountable set without rationals. [duplicate]

This question already has an answer here:

Problem: Find a closed, uncountable without rationals, subset in the unit interval.

Solution: Let X be any binary irrational number , replace the 1s with 2s in the decimal expansion .

Now we can use the digits of X as an index for a path through the cantor set, for example, if the first two digits of X are 0 2 then we take [0 , 1/3] and then we take [2/9 , 1/3] .

This subset is obviously closed and uncountable, and since we are using an irrational index all the elements of that subset will be irrational.

Ok, is there a big mistake here? And secondly, If all the end points of a cantor set are rational doesn't this contradict the above?

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• $[2,3]$ is closed, uncountable, and contains no rationals in the unit interval – mathworker21 Sep 3 '17 at 18:46
• What exactly is the subset in question - the intersection of the intervals you get from $X$? If so, that set has only one element. – Noah Schweber Sep 3 '17 at 18:52
• Answer: yes, and yes. With your procedure, you'll get lots of those rational points (they aren't all endpoints, $1/4$ is in the Cantor set). – Professor Vector Sep 3 '17 at 18:56
• The procedure is to pick a third and throw the rest, and then pick another third of that and throw the rest and so on. You pick the thirds based on the digits of X – Zee Sep 3 '17 at 18:59
• @Zee In that case you get only one point - e.g. the sequence $X=00000...$ will give you $[0,{1\over 3}]\cap [0,{1\over 9}]\cap [0,{1\over 27}]\cap ...=\{0\}.$ – Noah Schweber Sep 3 '17 at 19:00

I think what you want is the set of those $t \in [0,1]$ with base $3$ expansion $.t_1 t_2 t_3 \ldots$ (i.e. $t = \sum_{j=1}^\infty t_j 3^{-j}$) such that $t_j \in \{0,2\}$ if $x_j = 0$ and $t_j = 1$ if $x_j = 1$, where $X = .x_1 x_2 x_3 \ldots$ is the base $2$ expansion of the irrational number $X$. That would be a correct answer.

• That seems similar to what am saying, but doesn't this force the end point eventually to be irrational? Which is a contradiction – Zee Sep 3 '17 at 19:04
• @Zee : You could intersect $[0,1]$ with a subinterval having irrational endpoints, then keep only those $t$ in this new subinterval, if you are concerned about endpoints. Say, $t \in (\sqrt{2}-1, \sqrt{3}-1)$ with base $3$ expansion ... – Eric Towers Sep 3 '17 at 21:56
• What contradiction? You asked for a subset without rationals. – Robert Israel Sep 4 '17 at 0:38
• This answer is wrong. Let $X_n = 2^{-n-1/2}$ for each natural $n$. Then the base-2 expansion of $X_n$ starts with $n$ zeroes, and so there is a corresponding $t_n$ with base-3 expansion starting with $n$ zeros. Thus $t_n \to 0$ as $n \to \infty$, and hence the set constructed in this answer is not closed. – user21820 Sep 4 '17 at 1:59
• @user21820: It works if $X$ is a fixed irrational number used to construct the whole set. – Henning Makholm Sep 4 '17 at 4:12

Not that you asked about it, but you can avoid the Cantor set-ternary expansion bit by considering any $U\subset [0,1],$ $U$ open in $[0,1],$ containing $\mathbb Q \cap [0,1],$ with $m(U)<1/2.$ Then $[0,1]\setminus U$ is a closed subset of $[0,1]$ having measure greater than $1/2$ that contains no rational.

If you know about the regularity of Lebesgue measure, then let $E$ be the set of irrational numbers in $[0,1].$ We have $m(E)=1$ and by regularity,

$$m(E) = \sup \{ m(K): K \subset E, K \text { closed in } \mathbb R \}.$$

It follows that there are plenty of of closed subsets of $[0,1]$ having positive measure that contain only irrational numbers.

• Using measure here is massive overkill, though (and also doesn't help produce an example). – Noah Schweber Sep 3 '17 at 19:16
• Well measure theory is one of the tags, but thanks for your comment. – zhw. Sep 3 '17 at 19:19
• I didn't mean that negatively, I just meant to clarify for the OP what measure is and isn't doing in the context of this problem. – Noah Schweber Sep 3 '17 at 19:24
• Another nonconstructive example is as follows: Take the zero set of a Brownian motion $(B_t)$ . One may show that this is almost surely a perfect set with more than one point, hence uncountable. But $$P(\exists q \in \Bbb Q: B_q=0) \leq \sum_q P(B_q=0)=0$$ Thus we have our example, which is obviously overkill. – Shalop Sep 3 '17 at 23:59

This subset is obviously closed and uncountable, and since we are using an irrational index all the elements of that subset will be irrational.

The word "obviously" is one you should always be afraid of in math. While it is indeed obvious that your set is closed (being an intersection of closed sets) and it's not hard to show that it consists only of irrationals (although that isn't really trivial - you should say a bit about why it's true), it is not true that the set is uncountable!

I explained this in the case when $X=000...$ in the comments; your response was

that can't happen since $X$ is irrational.

However, it makes no difference what $X$ is; no matter what we pick for $X$, we'll always get a single point. In fact, it's easy to describe the point you'll get: it's just $X$ itself, but with each $1$ replaced by a $2$ and then interpreted as a ternary expansion. E.g. if we have $$X=0.010011000111...$$ your process will produce the set $$\{0.020022000222...\}.$$

(Don't believe me? What's a different point in the set you get from this $X$?)

To see why this is true, just note that at each step of the process you're determining another digit of the elements in the set you're building. E.g. with the $X$ above, by looking at the first three digits it's clear that in the set you build, every point's ternary expansion will begin "$0.020...$". So your whole process determines the point completely. The reason the Cantor set is uncountable is that you "keep having choices" - e.g. your first ternary digit could be $0$ or $2$, your second ternary digit could be $0$ or $2$, etc. By removing these choices as you've done, you get only one point.

• Mmmmm this feels weird though, doesn't this imply the cantor set degenerates into a bunch of single points eventually? – Zee Sep 3 '17 at 19:09
• @Zee No. It implies that the Cantor set is made up of single points (which is true of any set of points - that's what sets are). Each infinite binary sequence $X$ determines a single point of the Cantor set, and each point in the Cantor set corresponds to a unique infinite binary sequence. But the set of all infinite binary sequences is huge (by which I mean, it's uncountable). Each interval at each stage of the Cantor set construction corresponds to a finite binary string - e.g. $[{2\over 9},{1\over 3}]$ corresponds to the string "$01$" - (cont'd) – Noah Schweber Sep 3 '17 at 19:12
• and every infinite binary string beginning with this finite string will "name" a point in this interval (e.g. the sequences $0100000...$ and $01101010101....$ both name points in the interval $[{2\over 9}, {1\over 3}]$). – Noah Schweber Sep 3 '17 at 19:15
• @Zee Because there's two different notions of "double infinitely many times" (this is what my previous comment was getting at): one is to take the union of the stages, while the other is to take the infinite product of two-element sets. The first yields a countable set, while the second yields an uncountable set, and it's the second one that's going on here. E.g. the first is describing the set of all finite binary strings (length $1$ = double once, length $2$ = double twice, etc.) while the latter is describing the set of all infinite binary strings. This is vague, of course (cont'd) – Noah Schweber Sep 3 '17 at 21:01
• but so is what you're saying. The point is that using suggestive-but-imprecise language like "double infinitely many times" is a great way to mislead yourself. – Noah Schweber Sep 3 '17 at 21:03

The Cantor set $C \subset [0,1]$ is an uncountable disjoint union of sets $C_x, x \in C$ that are all homomorphic to $C$ (so perfect compact etc.). There can only be a rational in countably many of the $C_x$ so lots of them contain not rationals.

The union property is direct from $C \times C \simeq C$ which can be seen from the ternary expansion representation of the Cantor set: split the expansion of $x \in C$ into an odd and even indexed part to produce a pair of points in $C$. Merging two such expansion is the inverse.

An alternative but less elementary way to see the homeomorphism is to use Brouwer's theorem that all compact totally disconnected metric spaces without isolated points are homeomorphic to (each other and to )$C$.

• Are you certain that there are uncountably many copies of $C$ inside $C$? All the end points are rational, and each smaller copy of $C$ inside $C$ has an end point, right? – Arthur Sep 4 '17 at 6:19
• @Henno sir pliz help math.stackexchange.com/questions/3383987/… – jasmine Oct 7 at 10:35

Here is the simplest solution to your question. $\def\nn{\mathbb{N}} \def\less{\smallsetminus}$

Let $(r_n)_{n\in\nn}$ be a sequence enumerating all the rationals in $[0,1]$.

Let $S = [0,1] \less \bigcup_{n\in\nn} B(r_n,2^{-3-n})$, where $B(r,d)$ is the open interval centred at $r$ with radius $d$.

Then $S$ is clearly closed and has measure at least $1-\sum_{n=0}^\infty 2^{-2-n} = \frac12$.

• This gives an explicit construction for the same idea given in zhw.'s answer. – user21820 Sep 4 '17 at 3:42

I'm not sure if this is explicit enough for you:

Fix $0< \epsilon < 1$. Enumerate the rationals in $[0,1]$ by $\{q_n\}_{n \in \mathbb{N}}$. Around $q_n$ take the open ball $B(q_n, \epsilon 2^{-n})$.

Then $$[0,1] \setminus \bigcup_{n \in \mathbb{N}} B(q_n, \epsilon 2^{-n})$$ satsifies the properties you want.

It is evidently closed, and it must be uncountable, since we are removing (at most) a collection of intervals of total length $$\sum_{n=1}^\infty 2^{-n}\epsilon = \frac{\epsilon}{2} <1.$$

So then we have left a set of "length" (really, "measure") greater than $1-\displaystyle \frac\epsilon2$.

You should see that it is immediate (use the same idea I just gave) that a countable set can be covered by a countable collection of intervals of total length less than any positive number. Hence our set cannot be countable.

By construction, our set has no rationals.

Note: that we dealt with the rationals was inconsequential. The only property of the rationals that we needed for this construction was the fact that they are countable.