# Mathematics Prelims

## July 17, 2008

### The Cantor Set: An Uncountable Null Set

Filed under: Analysis,Measure Theory — cjohnson @ 12:29 am

We’ve seen that every countable set is null, which naturally leads to the question of whether or not any uncountable sets can be null.  The uncountable sets we use most often (namely intervals or unions of intervals) are not null, but constructing an uncountable null set known as the Cantor set is not very difficult.

The Cantor set is defined iteratively, with the actual set being the intersection of all iterations.  We begin by letting $C_0 = [0, 1]$.  Then we construct $C_1$ from $C_0$ by removing the middle third of $C_0$, this gives us $C_1 = [0, 1/3] \cup [2/3, 1]$.  We then construct $C_2$ by removing the middle third of the intervals in $C_1$, so we have $C_2 = [0, 1/9] \cup [2/9, 1/3] \cup [2/3, 7/9] \cup [8/9, 1]$.  In general, we will construct $C_n$ from $C_{n-1}$ by removing the middle third of any interval in $C_{n-1}$.  Finally, we define the Cantor set $C$ as follows.

$\displaystyle C = \bigcap_{n=0}^\infty C_n$

Now note that at any given iteration there are $2^n$ intervals (we start with $2^0 = 1$ intervals with $C_0$, then in each iteration double the number of intervals as we split each one into two pieces, the third on the left, and the third on the right).  Furthermore, each interval has length $\left( \frac{1}{3} \right)^n$.  This means the sum total length of $C_n$ is

$\displaystyle \sum_{n=1}^{2^n} \left( \frac{1}{3} \right)^n = \left( \frac{2}{3} \right)^n$

Let $\epsilon > 0$ be given.  Since $\left( \frac{2}{3} \right) < 1$, we can find an $n \in \mathbb{N}$ such that $\left( \frac{2}{3} \right)^n < \epsilon$, and so there exists an $n$ such that the sum total length of the intervals that compose $C_n$ is less than $\epsilon$. Now note that since $C$ is the intersection of sets that include $C_n$, the sum total length of $C$ must be less than that of $C_n$ which will be less than any given $\epsilon > 0$ for a large enough $n$.  This means that $C$ is null.

Now notice that the Cantor set is not empty, as the end points of any interval in some $C_n$ will be in every $C_n$. (Suppose you look at the end points of $C_n$.  Those points will be inside of intervals in each $C_m$ with $m < n$, and on the endpoints of intervals in each $C_m$ with $m \geq n$.)

We can show that the Cantor set is uncountable by showing that an element in the Cantor set is obtained if we take a number in [0, 1], write it in ternary, and that representation doesn’t have any ones.  Basically, if we were to write down the ternary representation of some number in [0, 1] we’d start off with the first place after the decimal point (the 1/3 spot).  If we write down a 0, we’re on the left-hand side of $C_0$, if we write down a 2 we’re on the right-hand side of $C_0$, and if we were to write down a 1, we’d be in the interval (1/3, 2/3) and thus not in the Cantor set.  When we write down the second digit after the decimal point (the 1/9 spot), we will “jump” to the left-hand interval of the interval we just split up if we write down a 0, to the right-hand interval if we write down a 2, and into the middle-third the interval we just split (i.e., outside of the Cantor set) if we write down a 1.  This is an injective mapping from the numbers in [0, 1] whose ternary expansion don’t have a one (which is an uncountable set) into $C$, and so the Cantor set must be uncountable.

## 5 Comments »

1. Absolutely great notes ty!

Comment by Batuhan — January 19, 2009 @ 7:28 pm

2. hello

i am doing a problem sheet and found this page unbelievably useful. Thank you so much, you made my day. i love maths so much. without it i would be a frog in a pond with no lily pads.

Comment by leah templeman — May 20, 2009 @ 11:34 am

3. Yeh, I like this, its very important to realise null sets maybe uncountable as I’ve just considered countable to be null.

One small thing I don’t get, why write C as an infinite intersection of sets, since C is just C_infinity. Or I guess… C_infinity isn’t intuitively apparent as you would have to apply the ‘removing middle third’ process an infinite amount of times. So the intersection notation is more logical.
But saying that, couldn’t one say C = lim[n->infinity] C_n ?

Comment by Colin — October 29, 2009 @ 12:00 am

4. Hi Colin,

You wrote, “But saying that, couldn’t one say C = lim[n->infinity] C_n ?”

I am wondering if you have in mind an interesting definition for the limit of a sequence of sets. I sat here and thought about it for about 5 minutes and realized there might be problems when trying to define a norm for sets that would work in this case. I would be interested in knowing if you (or anyone reading this) have a good set norm that would work here so we could actually talk about the limit of a sequence of sets.

Comment by Jeremy — January 2, 2011 @ 12:23 am

5. Thank you for this post it was really useful for a related problem.
However I’ve found a minor mistake:

At the total length of C_n the varible below the sum (I mean n = 1) shouldn’t be n, but sum other variable.

Comment by Argathron — April 5, 2011 @ 7:06 am

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