Hopefully we all remember Cantor as the guy to blame for ruining lives by delving too deeply into infinity. In addition to showing that some infinite sets are “larger” than others, we have a graphical representation of something called the Cantor set:

Not very impressive looking, is it? But it’s more interesting than it seems. For those who find it easier to work with numbers, imagine starting with the interval [0,1], and for others just imagine a perfectly straight wooden “ruler” 1 unit length long (inch, centimeter, whatever).

The Cantor Set is an iterated function system (IFS) which is a fancy way of saying that we take things one step at a time. We start with our interval [0,1] or 1 unit length, and we remove the middle third (as in the picture). Next, we remove the middle from the remaining two intervals/pieces of ruler (the intervals would be [0,1/3] & [2/3,1], in case you were wondering). We again take out the middle third, but now twice: we remove the middle third of each interval/piece of ruler. By the time we get to the fourth step, we have to remove 1/3 from intervals. We continue taking out the middle third of each remaining “level” of intervals/pieces of ruler, and the number of intervals at any level *n *from which we remove 1/3 is :

…and onward to infinity. Here’s the interesting part. Imagining that we really did this infinitely many times, how much have we removed? Well, at step *n *we remove 1/3 from remaining intervals/pieces of ruler, so we just let *n *go to infinity. We get this:

(the weird looking s-like thing is the uppercase Greek letter sigma and simply means “sum the following expression starting at n=1 and continue to infinity”).

I can almost here the groans and complains. I’ve forced people to confront algebra nightmares from high school for **this**? No. The interesting thing is how much is left. Think about it: every time we removed **the middle** of some number of remaining intervals/pieces of ruler. We remove the middle of intervals/pieces of ruler infinitely many times, each time leaving TWICE as much as we took away. So we started out with (a unit) 1, we removed a total amount/length 1, and we wind up with twice as much as we began with.

Why doesn’t my bank account work that way?

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Because your bank knows how to count 🙂

No. Sure, if they knew how they’d know the fallacy in what I’ve written. But alas, they’re the broken clock that’s right twice a millennium. Here they’re just lucky, not familiar with measure theory and why Cantor’s set is uncountable with measure 0. Or maybe they are and I should switch banks…

The bank would argue that if you withdraw 1/3 of your balance, then 1/3 of what is left, and keep going forever like this, you will end up with no money in your account at the end of your infinite sequence of withdrawals. I will let the other readers find a whole in your faulty argument.

I’m not simply withdrawing 3rds, though. I’m withdrawing lengths/amounts/intervals that are always and only ever MIDDLE thirds, which means every time I remove an amount/interval, I leave behind the endpoints that will always remain even if I continue to remove middle thirds infinitely many times.

The flaw isn’t in the counterintuitive result of removing infinite intervals equal to the total original interval and having infinitely many points left, it’s in the way that points are described. The cantor set (the points left over after infinitely many 3rds are removed) is interesting because, among other things, it is an uncountably infinite set (as opposed to countably infinite like the rationals, integers, etc.) with measure 0.

Does not make any difference to a banker who is interested only in the amount of money ( = total length of the intervals left).

In certain cases, where banks allow units less than a cent (even much less) to count as an amount, than the amount that matters is the amount removed and left. Intervals aren’t the only thing that matter, and they don’t matter to bankers while amounts do (intervals as amounts do to). If I have $100 in the bank and I remove amounts equal to middle thirds as in the main example such that I am always leaving a certain amount of money behind, then the bank’s computer might have a problem as I would not have removed the entire amount in my account, but will have removed $100. Of course, it would take a while to remove this $100. Specifically, it would take forever.

But if the bank rounds the balance to a penny, it will not take forever, because you balance after n transactions would be (2/3)^n and it is less than half a penny for n logarithmic in the original balance, so your balance will become 0 pretty fast.

Clearly. And in an ideal world, I’d be posting how fascinating the Cantor set is by describing it as an uncountably infinite set with measure 0 with a complement of measure 1 (in the “standard” ternary version) followed by the Cantor function. However, if I restricted myself to discussions at the level of discourse I generally employ, not only would nobody visit this blog (which wouldn’t bother me) but it would defeat its purpose (to the extent one exists).

Should be b(2/3)^n where b is the original balance in the previous comment.

….. should be a hole, not a whole in my reply, strange typo….