Why is the middle third Cantor set written as this?

My first question is, is the middle third Cantor set the same as the Cantor set? I've never heard it called the middle third Cantor set.

Secondly, why is this true:

"I’m going to assume that Cantor set here refers to the standard middle-thirds Cantor set $C$ described here. It can be described as the set of real numbers in $[0,1]$ having ternary expansions using only the digits $0$ and $2$, i.e., real numbers of the form $$\sum_{n=1}^\infty \frac{a_n}{3^n},$$ where each $a_n$ is either $0$ or $2$. "

I'm not truly understanding why it would be 0 or 2 for $a_n$. I understand that the numbers from [0,1] can be written in ternary as 0,1, or 2. I don't understand why the "1" is taken out. Is it because we take the middle third out every time and we lose anything that can be written in ternary as "1"?

It's called the middle-thirds Cantor set because in general you can construct a class of sets with similar properties using a similar but scaled construction. For example, you can start with $[0,1]$, remove an interval of length $\frac{1}{2}$ from the center. Then you have $2$ intervals of length $\frac{1}{4}$, remove an interval of length $\frac{1}{8}$ from the center. Inductively, at each step you have a disjoint union of intervals of length $l$ remaining in your set, and you remove an interval of length $l/2$ from each interval. This set, which we might call the Cantor "middle-halves" set, has many of the same properties of the Cantor middle-thirds set. From this you can imagine constructing the "middle-fourths" set and many other Cantor-type sets. The middle-thirds set is sort of standard because it's the easiest Cantor-type set to construct (mainly it's easy to figure out the lengths of the intervals at each step).

The reason for the ternary expansion is precisely as you stated, the middle-thirds construction removes any numbers with absolutely necessary $1$-s in the ternary expansion. You should check this for yourself for a few cases: for example, check if $0.1abcd...$ (ternary) can be in the middle-thirds set, then $0.01abcd...$, $0.21abcd...$, and so on. Then you'll see why the Cantor set construction removes them. This excludes cases like $1/3$, which lies in the Cantor set and has ternary expansion $0.1000...$, because it can also be written $0.0222...$ .

• Correct me if I'm wrong, but isn't 0.01 = 1/9 in our normal numerical system, and isn't that left when you take away the first middle third? – H5159 Nov 22 '14 at 2:22
• Yes, that's precisely my point. – Gyu Eun Lee Nov 22 '14 at 2:23
• But if 1/9 is left doesn't that mean our summation includes ternary values with 1 in them? – H5159 Nov 22 '14 at 2:25
• Ah, I see. The problem is that ternary expansions are not unique. $1/9$ can be written as $0.01000...$, but it can also be written as $0.00222...$. (Like how $1 = 0.999...$.) The convention is to write the ternary expansion with only $0$-s and $2$-s if possible, and throw out numbers that require a $1$ no matter what you do. Notice that you also keep $1/3$ (and in general lots of $k/3^n$-s) by the middle-thirds construction (they lie at the endpoints of intervals you don't delete), but the ones you keep can all be written without $1$-s. – Gyu Eun Lee Nov 22 '14 at 2:34
• Not quite; for example, $2/3$ is an endpoint we keep as well (but of course $2/3 = 0.2000...$ so this is fine). We lose any values that must have $1$ in its ternary expansion, not the ones that merely can be written with $1$. The numbers $1/3^n$ are examples of numbers that can be written with $1$ in ternary, but they don't have to be written with $1$. You can always do $0.000....0222...$ (0-s until the $n+1$-th ternary place) for these instead of $0.000...01000...$ ($0$-s until the $n$-th ternary place). Therefore we keep all $1/3^n$, though this doesn't exhaust the list of endpoints. – Gyu Eun Lee Nov 22 '14 at 2:50

You can build several "essentially" different cantor sets, some have Lebesgue measure 0, some have positive measure. For instance, take the unit interval, remove the "middle fifth" (take a linear function to the interval $[0, 5]$ and remove the preimage of the open $(2,3)$). On step 2, remove the "middle twenty-fifth" of each of the two remaining intervals, etc. At the end you will get a nowhere dense closed set with no isolated points (so, a Cantor set), but its measure will be strictly positive.

The standard Cantor is cbuild by removing one third of each interval at each step, and yes, it relates to the '1' digit in the ternary expansion of the number.

First question: typically the Cantor set is the middle third Cantor set, while a Cantor set can be any of a number of similar sets, including variants that have positive measure.

Second question: essentially, yes. Being more precise, in the $i$th stage of the construction one removes all the remaining numbers whose ternary expansion has a $1$ in the $i$th place. Hence at the end of the construction, all the numbers have no $1$s in their ternary expansion.