Related to a question I asked earlier.:

Let $F$ be the subset of $[0,1]$ constructed in the same manner as the Cantor set except that each of the intervals removed at the $n$th iteration has length $\frac{\alpha}{3^{n}}$ with $0 < \alpha < 1$.

I need to show that $F$'s complement, $[0,1]\backslash F$ is dense in $[0,1]$.

I know that a $[0,1]\backslash F$ is dense in $[0,1]$ if any open interval in $[0,1]$ contains a point of $[0,1]\backslash F$.

Obviously, the open intervals in $[0,1]$ comprising $[0,1]\backslash F$ contain points $[0,1]\backslash F$,so, I just need to worry about the open intervals inside $[0,1]$ that are not part of $[0,1]\backslash F$.

However, those would be open intervals inside the Cantor set, and while I know that the Cantor set does not contain any intervals of positive measure, I don't know how to show this all epsilon-y (i.e., with neighborhoods, ets). Therefore, I was asking how to formally prove this (either an outline or a full proof is fine; hints are fine, too, if you don't mind follow-up questions.)

  • $\begingroup$ @Arthur, so total length remaining is $1-\alpha$, so a length of $\alpha$ is removed. But, I've got to tell you, I am having a lot of trouble figuring out which points are guaranteed to be removed. I guess it would be right endpoint of closed interval $+ \epsilon$ and left endpoint of closed interval $-\epsilon$, but because of the $\alpha$'s, as opposed to even $1/3$rds, I am really having trouble visualizing this. Please elaborate more on what you mean. If you turned your comment into an answer with appropriate details, I might be willing to accept your answer and give you points. $\endgroup$ – ALannister Oct 22 '15 at 13:25
  • $\begingroup$ You're right, I was a bit quick making that comment. However, if the complement is not dense in $[0, 1]$, then there must be an open interval somewhere in $F$. That would soon get you into a contradiction. $\endgroup$ – Arthur Oct 22 '15 at 13:26
  • $\begingroup$ @Arthur, again, because of the $\alpha$, it is not $1$ like it is with the usual Cantor set, I'm not sure how to write out that contradiction. I have an idea that suppose $[0,1]\backslash F$ is not dense in $[0,1]$. Then, $F$ must contain some open interval $(a,b)$ (does it need to be a NONEMPTY open interval? Seems it would defeat the purpose w/o this condition). Suppose that $(a,b)$ is nonempty (i.e., where $a \neq b$). Then, choose $n \in \mathbb{N}$ such that $\frac{\alpha}{3^{n}}<b-a$. Since $F$ is contained in a countable intersection of closed intervals, all of length $< (b-a)$... $\endgroup$ – ALannister Oct 22 '15 at 13:46
  • $\begingroup$ @Arthur (continued from above), and on each iteration, the lengths of the intervals decreases, this intersection cannot contain $(a,b)$. Is that correct? If not, please let me know specifically what to change about it. $\endgroup$ – ALannister Oct 22 '15 at 13:48
  • $\begingroup$ First of all, the total "length" (the measure) of $F$ isn't necessarily $0$. The total length of all the removed intervals is actually $\alpha$, so $1-\alpha$ is left in $F$. But at each iteration, every component is divided into two equal parts (and then some is removed), so for each iteration, the length of the longest interval contained in the set becomes less than half of what it was. From there it's easy to see that there cannot be any positive-length intervals left in $F$. $\endgroup$ – Arthur Oct 22 '15 at 14:09

Let $F_i$ be the result of the $i$-th iteration of removing all the $\alpha/3^i$ wide open intervals. That means $F_0 = [0,1]$, $F_1 = [0,\frac12-\frac\alpha6]\cup [\frac12 + \frac\alpha6, 1]$ and so on. We then have $F = \bigcap_i F_i$.

Now, every $F_i$ consists of $2^i$ disjoint, closed intervals of equal length. That means that each such closed interval is less than $1/2^i$ wide.

Assume $(a, b) \subseteq F$ with $a < b$. That means that $(a, b) \subseteq F_i$ for all $i$, by definition of $\bigcap$. But that can't be true, since $b-a > 1/2^k$ for some $k$, and in the corresponding $F_k$ there is no room for an interval of that width.

Therefore there can be no non-empty open interval in $F$, which makes the complement of $F$ dense.

  • $\begingroup$ Why not just assuming F is dense, then it must intersect every open interval inside [0,1], then use the interval $(5\alpha/12,6\alpha/12)$ as an open interval which does not intersect $F_1$ and consequently does not intersect F? That is, an open interval which is inside the middle removed set at the first step. That will show that F is not dense, Hence [0,1]\F is dense. $\endgroup$ – Richard Clare Oct 11 '18 at 13:50
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    $\begingroup$ @RichardClare A set can be dense and have a dense complement (the rationals in the reals, for instance). A set can be nondense and have nondense complement (the positive reals in the reals, for instance). So it doesn't help to show that $F$ isn't dense. Showing that $F$ is nowhere dense, on the other hand, that would help. But that takes a bit more work. $\endgroup$ – Arthur Oct 11 '18 at 13:55
  • $\begingroup$ Arthur, but [0,1] is dense, and is the union of F and F complement is [0,1] so if F isn’t dense, F complement must be ?!? $\endgroup$ – Richard Clare Oct 11 '18 at 14:08
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    $\begingroup$ @RichardClare The exact same can be said about $F_1$ instead of $F$, but the complement of $F_1$ isn't dense in $[0,1]$. Just because a set isn't dense, doesn't mean its complement must be. $\endgroup$ – Arthur Oct 11 '18 at 14:10
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    $\begingroup$ @RichardClare No. I've proven that the complement of $F$ is dense, which is a weaker statement than $F$ being nowhere dense. I did this by showing that every non-empty interval contains points from said complement, which is the definition of "dense". $\endgroup$ – Arthur Oct 11 '18 at 14:17

Hint: The length of any of the closed intervals not removed by step $n$ have length at most $2^{-n}$.

  • $\begingroup$ Any of the individual closed intervals? Or the complete amount removed??? I'm still so confused...And how would doing that show me that no open interval is in there??? $\endgroup$ – ALannister Oct 22 '15 at 15:59
  • $\begingroup$ anyway, are you sure you don't mean $3^{-n}$ $\endgroup$ – ALannister Oct 22 '15 at 16:08
  • $\begingroup$ What you are trying to show is that the union of all the open intervals removed is dense. What does it say if the maximum distance between any two open intervals is at most $2^{-n}$? $\endgroup$ – robjohn Oct 22 '15 at 16:09
  • $\begingroup$ I don't know. I seriously don't. $\endgroup$ – ALannister Oct 22 '15 at 16:10
  • $\begingroup$ Remember that you are dividing each closed interval into two closed interval by adding an open interval in the middle.. $\endgroup$ – robjohn Oct 22 '15 at 16:10

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