4
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We construct a subset of $[0, 1]$ by successively removing middle half of each remaining intervals: starting with $I_0 = [0, 1]$, we let $I_1 = [0,\frac{1}{4}] ∪[\frac{3}{4}, 1]$, and then $$ I_2 = [0,\frac{1}{16}] ∪ [\frac{3}{16},\frac{4}{16}]∪[\frac{12}{16},\frac{13}{16}]∪[\frac{15}{16},\frac{16}{16}]$$

Now, let $L$ be the set remaining as a result of limit of this process; that is $$ L = I_0 ∩ I_1 ∩ I_2 ∩ ... ∩ \ I_n ∩ ...$$

Show that the limit of this process takes away total length of exactly $1$ unit, and leaves behind a set, $L$, that is still uncountable, of cardinality $c$.

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  • 1
    $\begingroup$ What is the length of the intervals you remove at the $n$th step? If you add them all you will get the first part of your question. As a hint, look in wikipedia for the "Cantor set" (in your case you remove a bigger part of the intervals, but essentially this does not change anything). $\endgroup$ – Dim A Apr 8 '17 at 14:23
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Hint: Try several iterations, and render the bounding values (which must be retained forevermore in subsequent iterations) in base 4 after each step. What pattern emerges? Then, how do you show that the cardinality of the final surviving values is $2^{\aleph_0}$?

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