My question is after the construction of the Cantor set which you might want to skip over that part. This question is not philosophical at all it is pertinent to my understanding of measure and space. Please If you think this is absurd, don't bother replying.
Construction of the Cantor Set:
We will denote the cantor set as $\mathfrak{C}\subset\mathbb{I}$
, $\mathbb{I}=[0,1]$
as defined in section 1, as mentioned above we construct $\mathfrak{C}$
by removing middle third of the set (not including its endpoints). We denote this first slice in $\mathbb{I}$
as $$\mathfrak{c_{1}}=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3},1\right]=\mathbb{I}\setminus\left(\frac{1}{3},\frac{2}{3}\right)$$
Next we cut out the middle third of each interval in $\mathfrak{c_{1}}$
, that is the intervals $\left[0,\frac{1}{3}\right]\text{ and }\left[\frac{2}{3},1\right]$
so we get
$$\left[0,\frac{1}{3}\right]\to\left[\frac{0}{9},\frac{1}{9}\right]\cup\left[\frac{2}{9},\frac{3}{9}\right]=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9},\frac{1}{3}\right]$$
$$\left[\frac{2}{3},1\right]\to\left[\frac{6}{9},\frac{7}{9}\right]\cup\left[\frac{8}{9},\frac{9}{9}\right]=\left[\frac{2}{3},\frac{7}{9}\right]\cup\left[\frac{8}{9},1\right]$$
So we write $$\mathfrak{c_{2}}=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9},\frac{1}{3}\right]\cup\left[\frac{2}{3},\frac{7}{9}\right]\cup\left[\frac{8}{9},1\right]$$
Hence we have that $\mathfrak{c_{2}}$
is the union of $2^{2}$
intervals all of which are of the form $[k/3^{2},(k+1)/3^{2}]$
, if we continue this process for all $n\in\mathbb{N}$
we find that each $\mathfrak{c}_{n}$
is constructed of $2^{n}$
intervals all of the form $[k/3^{n},(k+1)/3^{n}]$
, and $\mathfrak{c}_{n+1}$
will be obtained by taking the middle third out of each interval in $\mathfrak{c}_{n}$
. The cantor set is what remain after we continue this processv for all $n\in\mathbb{N}$
and then take the intersection over every $\mathfrak{c}_{i}$
, or we could write $$\lim_{n\to\infty}\bigcap_{i=1}^{n}\mathfrak{c}_{i}=\mathfrak{C}$$
Where $\mathfrak{C}$
is the Cantor set.
My Question:
Why do we say that the Cantor set has no length? I get that the length of $\mathfrak{c_1}=\left(\frac{2}{3}\right)^1$ and that the length of $\mathfrak{c}_n=\left(\frac{2}{3}\right)^n$ so
$$\lim_{n\to \infty} \left(\frac{2}{3}\right)^n= 0$$
Since, $\mathfrak{C}\subseteq\mathfrak{c}_n$ if follows that $\mathfrak{C}$ has a length of zero. Or, even you could argue that
$$\lim_{n\to\infty}\frac{k}{3^{n}}=\lim_{n\to\infty}\frac{k+1}{3^{n}}=0$$
Since, the endpoints of each interval in $\mathfrak{c}_n$ converge to the same point each interval will become a single point in interval $\mathbb{I}$, and a single point $x\in\mathbb{R}$ has a measure of zero, i.e. $d(x,x)=0$. So, summing up distances of each intervals (of which there must be $2^{\aleph_0}$ many) in $\mathfrak{C}$, you get that the distance is zero.
It is the concept of the limit and space that is tripping me up. Even if $n\to \infty$ $\frac{k}{3^n}\neq 0$ unless $k=0$ (in which case $\frac{k+1}{3^n}\neq 0$). In my mind $\frac{k}{3^n}$ can never equal $\frac{k+1}{3^n}$ because $k\in\mathbb{N}$ and $k+1$ is $k$'s successor (thus they are clearly not equal), so in a world of only the natural numbers it is clear that $k$ and $k+1$ don't occupy the same space. However, in $\mathbb{R}$ if we take the limit $n \to \infty$ of both $\frac{k}{3^n}$ and $\frac{k+1}{3^n}$ they end up equaling the same point in space, even though there numerators are different. Clearly, a difference of $1$ is negligible in respects to $\infty$. But in my mind the every point in $\mathbb{R}^n$ is distinct, if you changed the trillionth decimal place in $\sqrt{2}$ and left every other digit intact, this new number would not share the same space (point) as $\sqrt{2}$ in $\mathbb{R}$. And in this same vein I feel like $\frac{k}{3^n}$ and $\frac{k+1}{3^n}$ will never be equal, they will differ by some decimal. And ultimately that would mean that the Cantor set has a distance/measure.
Can someone explain to me why we would think differently about this? I am taking a first course in Analysis but please feel free to go beyond this If it will give a more rigorous explanation for why the cantor set has no distance. (Also, point out any abuses in language that I might have made.) Thanks!
Best Answer
I propose you trying to work this out with the following approach, which uses pretty elementary properties of Borel measure on the real line:
To form the Cantor set you "removed":
$$1\;\;\text{interval of length}\;\;\frac13\\2\;\;\text{intervals of length}\;\;\frac19\\\ldots$$
All in all, you removed disjoint intervals of total length
$$\sum_{n=1}^\infty\frac{2^{n-1}}{3^n}=\frac12\frac{\frac23}{1-\frac23}=1$$
Since everything's happening in the unit interval, what is left (properties!) of it after the above removal, i.e. Cantor's set, has length $\;1-1=0\;$