# Why does one third belong to the Cantor set?

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?

• 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. Jul 8, 2015 at 20:57

$(0.1)_3 = (0.0222\ldots)_3$, just as $0.1 = 0.0999\ldots$ in base $10$.

• lol you win by 4 seconds. =) Jul 8, 2015 at 7:03
• @user21820 Apparently you typed too many $2$s :) Jul 8, 2015 at 21:58
• @HagenvonEitzen: Yea.. all because I wanted to make an infinitesimal attempt to emphasize the infinite ternary expansion. Jul 9, 2015 at 13:44

$0.02222222222222222222..._3 = 0.1_3$

• Either you wrote too many $2$s or infinitely too few $2$s. I can't decide. Jul 8, 2015 at 7:12
• @columbus8myhw: Hahaha.. I held down the 2 on my keyboard for about one second. Jul 8, 2015 at 7:15
• @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). Jul 8, 2015 at 8:17
• @CiaPan: Haha columbus8myhw was joking.. Jul 8, 2015 at 8:19
• @Brian Another way would be 7.
– Paul
Jul 8, 2015 at 18:35

Any endpoint of removed intervals belongs to Cantor set.