We know that all numbers that belong to the Cantor set have a ternary representation with only 0's and 2's but, for example, $\frac{1}{3}=(0.1)_3$ and $\frac{1}{3}$ belong to the Cantor set. I don't understand and I have already read tons of sources. Can somebody explain to me what is going on?
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1$\begingroup$ The answers below are, of course correct. Perhaps another way to go is to realize that we only delete open intervals in the usual iterated construction. So, first we delete $(1/3,2/3)$ which leaves 1/3 in the set and it never gets subsequently deleted. $\endgroup$– Eric TowersJul 8, 2015 at 20:57
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3 Answers
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$(0.1)_3 = (0.0222\ldots)_3$, just as $0.1 = 0.0999\ldots$ in base $10$.
answered Jul 8, 2015 at 7:03
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6$\begingroup$ @user21820 Apparently you typed too many $2$s :) $\endgroup$ Jul 8, 2015 at 21:58
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$\begingroup$ @HagenvonEitzen: Yea.. all because I wanted to make an infinitesimal attempt to emphasize the infinite ternary expansion. $\endgroup$ Jul 9, 2015 at 13:44
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$0.02222222222222222222..._3 = 0.1_3$
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24$\begingroup$ Either you wrote too many $2$s or infinitely too few $2$s. I can't decide. $\endgroup$ Jul 8, 2015 at 7:12
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3$\begingroup$ @columbus8myhw: Hahaha.. I held down the
2on my keyboard for about one second. $\endgroup$ Jul 8, 2015 at 7:15 -
2$\begingroup$ @columbus8myhw No, it's not too many. The notation means there is 18 twos, and then a two infinitely repeating. That's perfectly true (despite other notations, like $0.022222222\dots_3$ or $0.0222\dots_3$ or $0.022\dots_3$, which result in the same meaning). $\endgroup$– CiaPanJul 8, 2015 at 8:17
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Any endpoint of removed intervals belongs to Cantor set.
