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I have known that there exists some Cantor-like sets with positive measure, and wondering if I can prove that any any Cantor-like set is uncountable by construting or just proving the existence of a bijection of any two Cantor-like sets.
Maybe there are several different proofs of this property, but I want to know will this method work.
A Cantor-like set is a set construting in the following way:
Removing $2^{k-1}$ centrally situated open intervals of length $l_k$ at each $k^{th}$ stage, with $$l_1+2l_2+...+2^{k-1}l_k<1$$

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    $\begingroup$ What is your definition of a cantor like set $\endgroup$ – TomGrubb Sep 30 '18 at 14:55
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    $\begingroup$ If you're considering fat Cantor sets, then they can't be countable, because they have positive Lebesgue measure. $\endgroup$ – Matematleta Sep 30 '18 at 15:09
  • $\begingroup$ @Driver8 You should write an official answer even if it is very short. $\endgroup$ – Paul Frost Sep 30 '18 at 15:27
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    $\begingroup$ As @TomGrubb has suggested, you need to tell us what definition of "Cantor set" you want to make use of. This is because there are several properties of Cantor sets that imply being uncountable, and some of these properties might be part of your definition of "Cantor set" (and thus an answer based on such a property would work for you) and other of these properties might not be part of your definition of "Cantor set" (and thus an answer based on such a property would not work for you unless accompanied by a proof that your definition of "Cantor set" implies the property). $\endgroup$ – Dave L. Renfro Sep 30 '18 at 16:03
  • $\begingroup$ @Matematleta What I want to know is if I can construt a bijection of a Cantor set to a fat Cantor set as you called, which gives a alternative proof. $\endgroup$ – Hugo Oct 1 '18 at 5:45
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If you take two sets which are constructed by removing centrally situated open intervals you have a for each iteration a homeomorphism between the two constructions, e.g. a piecewise linear map defined on the partition of [0,1]. These homeomorphism have a limit function which provides a bijection on the two Cantor-like sets.

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  • $\begingroup$ Why is the limit function also injective? $\endgroup$ – Hugo Oct 3 '18 at 12:08
  • $\begingroup$ If the limit set doesn't contain an interval you'll find a removed interval between two arbitrary points. On these intervals the limit function is striktly monoton and therefore striktly monoton everywhere. $\endgroup$ – TomTom314 Oct 3 '18 at 13:55
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If you're considering fat Cantor sets, then they can't be countable, because they have positive Lebesgue measure.

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  • $\begingroup$ Of course you are right, but what about any "thin" Cantor sets? Is there an easy way to give a bijection of them. $\endgroup$ – Hugo Oct 1 '18 at 5:59
  • $\begingroup$ All Cantor sets are unconuntable, so they have cardinality $\frak c$. Do you want to construct bijections? $\endgroup$ – Matematleta Oct 1 '18 at 15:29
  • $\begingroup$ It is a fact that all Cantor sets are uncountable, there are many approaches to this consequence. However, a proof is often given by analyzing the properties of the sets. But there may be an approach without doing such an analysis. In particular, if there is an easy way to construct a bijection between any Cantor set, then we can deduce that any Cantor set is uncountable since there exist some Cantor sets, which are uncountable. $\endgroup$ – Hugo Oct 2 '18 at 3:00

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