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Suppose we have $\left[1-\frac{1}{x},1+\frac{1}{x}\right]$. As $x\to\infty$, the length of the interval approaches $[1,1]$; however, it seems this is the same as the singleton $\{1\}$ and the interval is always uncountable as $x$ approaches infinity.

Does the cantor set contain countably infinite intervals of zero length? Yet the Cantor set is uncountable so I reason the intervals of zero length are uncountable. This includes my example.

Is my interval countable or uncountable? Explain why?

Somos
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Arbuja
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    $[1-\frac1x,1+\frac1x]$ is uncountable for all real $x$, though the limit as $x\to\infty$ is a singleton – J. W. Tanner Apr 22 '20 at 17:05
  • I wish there was some way I can visualize the uncountability of the cantor set. – Arbuja Apr 22 '20 at 17:06
  • @Arbuja See my answer. The points in the Cantor set have a simple description in terms of their base-$3$ expansion. – Noah Schweber Apr 22 '20 at 17:10
  • At the first stage, we kill the middle third. So any element in the set will be on the left or on the right. Say we go left. Then we kill the middle third again, and any element must be on the left or the right of this piece. And so on. In fact, every point in the Cantor set corresponds to an infinite list of left/right choices. But then there's one point for every function $\mathb N \to 2$, and there are uncountably many such functions. – HallaSurvivor Apr 22 '20 at 17:11
  • Your first paragraph does not make sense. There is no definition of a sequence of sets approaching a "limit". The intersection of all of the sets is the singleton ${1}$ but this is not a limit. The length of the intervals does approach $0$. – Somos Apr 22 '20 at 17:12
  • See here for a number of ways to see the Cantor set is uncountable. – HallaSurvivor Apr 22 '20 at 17:12
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    @Somos: There is no definition of a sequence of sets approaching a "limit". --- Actually, there are many such (non-equivalent) definitions. The most common is this notion, which is widely used in measure theory, probability, and some other fields. Another that is well known is convergence in the sense of the Hausdorff metric. Another is convergence in the sense of Kuratowski. – Dave L. Renfro Apr 22 '20 at 17:50

2 Answers2

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A (closed) "interval of zero length" is just a single point, hence very countable. This isn't a contradiction, since "cardinality isn't continuous:" we don't have $$\vert \bigcap_{i\in\mathbb{N}} A_i\vert=\lim_{i\rightarrow\infty} \vert \bigcap_{j<i}A_i\vert.$$ (Note that by contrast, there is a reasonable sense in which measure is "continuous," at least for non-pathological sets: namely, countable additivity.)

As to the Cantor set, it in fact has uncountably many "components." The obvious ones are the endpoints of the "stages" of its construction, and there are indeed only countably many of those, but there are also lots of other points - namely, the points in the Cantor set are exactly those points in $[0,1]$ whose base-$3$ expansion doesn't contain any $1$s, but most of these aren't endpoints in any of the "stages." (It's a good exercise to figure out an exact characterization of the ternary expansions of "non-endpointy" elements of the Cantor set.)

Noah Schweber
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We usually take $[a,a] = \{ a \}$.

For your first question, this is consistent because

$$\lim [1-1/n, 1+1/n] = \bigcap [1-1/n, 1+1/n] = \{1\}$$

For your second, the Cantor set is actually uncountable. So it is the uncountable union of singletons (which one really shouldn't call intervals of 0 length), and indeed contains no intervals (of positive length, if you must).

This should provide a useful change of intuition. Lengths (formally, measures) and cardinalities do not need to be preserved under limits of sets. Though there is a lot of interesting mathematics explaining when they are preserved!

I hope this helps ^_^

HallaSurvivor
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