## Double for Nothing, part 2

I know it’s been a while1 since the last post, but we’re not quite dead yet…

I’ll explain why it’s been so long at the very end of the post, but in the meantime, we’ve got some math to explore!

The Banach-Tarski paradox says that you can take a ball, cut it up into a handful of pieces, then rearrange them in order to get two balls identical to the original.

Impossible, right?

Wrong!2

It certainly seems impossible, though. After all, if all you do is cut up the ball, and move the pieces around (no stretching required!) then the volume of stuff from the ball shouldn’t change.

But you can duplicate a ball! The trick is that rotations can create points, seemingly out of nowhere.

Here’s a simple example: Take a circle, with a single point missing.

How can we fill back in that hole? The obvious thing to do is just infinitesimally stretch the circle to fill in that single-point gap.

But let’s rule out stretching — we know that stretching things mathematically can create points. Can we do it with just a rotation?

If we pick just the right set of points to rotate, we can!

Let’s start by taking the point one radian3 clockwise of the original point. If we rotate it counterclockwise around the circle, it’ll fill the gap we had.

So, of course, that point should be in the set we will rotate. Of course, moving that point leaves another gap, so we need to also rotate the point one radian clockwise of that. And then we need another point to fill in that gap…

…and so on, and so forth.

The trick is that we picked a special angle. Recall that a circle has 360 degrees, or, equivalently, $2\pi$ radians. If we keep picking points one radian clockwise of the original gap, we go around the circle once, then twice, then more, but we will never end up back where we started. That is because $2\pi$ is an irrational number! (Details in this footnote:4)

The original hole is filled by the point that was 1 radian away. The hole left at 1 radian away is filled by the point two radians away. That hole at 2 radians is filled by the point at 3 radians… and so on, so a billion radians later, the hole left by the point a billion radians away is filled in by the point a billion and one radians away.

Sure, by this point, we’ve wrapped around the circle, oooohh, say, 160 million times, but we have never repeated a point! All thanks to $2\pi$ being irrational.

So, why isn’t there a hole at the end of all this? Well, there is no end. We’re kind of pulling a point out of infinity to fill the gap.5

Of course, creating a single point is not so impressive. So, let’s get back to the Banach-Tarski paradox.

As we talked about in the last post, the key trick is not really about geometry at all. I’m going to review some of what we discussed last time, but if you haven’t read or don’t remember Double for Nothing: the Banach-Tarski Paradox, you probably should before finishing this post.

If we take a ball, we can rotate it in different directions, forward (F), backward (B), right (R), and left (L). And, we could do multiple rotations in a row, for example, FRB would be backwards rotation, right rotation, then forward rotation.6

We can put all of these “words” representing rotations into a graph, where, for each letter in the word, F means you go up, L left, etc.. The center point, which represents no rotations, we can label N.

Thus, a series of rotations is represented by a word which is represented by a point on this branched graph. Of course, we can do any length of words (i.e., any number of rotations), so we get an infinite graph.

The “words” starting with L represent the rotations ending with a left rotation, and are the ones on the left side of the graph. (Again, words are series of rotations are points on this graph.)

The key observation from last time was that if we take the “words” starting with L, then undo the last rotation by rotating right, we end up with all the words except the ones on the right!

Where do all these extra points come from? Well, like the circle example from earlier, in some sense we’re “pulling them from infinity,” i.e., pulling them out of those infinitely small branches down in the graph.

The key of the Banach-Tarski paradox is figuring out how to get this “creation” of points on the graph to work on a ball instead.

To do that, we need to associate points in the ball with this graph somehow. Fortunately, the basic idea is not too hard. The points on the graph are supposed to represent words which represent series of rotations of a ball. Thus, we’ll try to associate each point on the graph, i.e., a word, with points on the sphere that we find via those rotations.

Grab the ball. The word that represented no rotations, N, we’ll associate with the “north pole” of the ball. The north pole, though, is just a point on the surface, and we want to duplicate the entire ball. So, let N actually represent all of the points below the north pole through the inside of the ball, all the way down to the core. (Though not including the center point at the core itself.) Thus, N represents a little line segment.

For every other series of rotations, after you rotate, the word (i.e., point on the graph) representing those rotations will represent the line from the new north pole to the center of the ball. For example, for the rotation L, you would rotate left, and that new north pole and the points under it are now “L.”

To make this all work, it’s very important that two different words, i.e. series of rotations, don’t represent the same set of points. To guarantee that, we need to pick the angle we rotate carefully. Fortunately, like in the circle rotation example earlier, it’s not too hard. One traditional angle is $\arccos(1/3)$, but there are infinitely many angles that would work.7 If we pick that angle, each word, or series of rotations, will rotate a new point to the north pole.

Great! We’ve taken the ball and identified its points with the words, which are points on the branched graph. And those points are spread out all over the ball — it turns out you can get points spread out evenly all over the ball with an arbitrary number of rotations.

Except…

Well, we’ve actually missed almost all the points of the ball!

Even though our words represent points that are evenly spread out all of the ball, so that it would look like we’ve covered everything, we’re actually missing “most” of the points in the ball!8

Fortunately, there’s an easy way to fix this.

Simply pick one of the points we missed on our first go, and start again with that point as the new north pole. We can associate the word N (for no rotations) with both this north pole and the original one we picked. Then, we can do all the rotations, like before, and associate their words with the new points as well. After doing this, we’ll have two line segments of points from the surface to the middle of the ball for each “word” like N or FR or BBBR, but we’ll lump them together into one set of points.

Unfortunately, we’re still missing most of the points in the ball. So, we pick yet another new north pole from the leftover points, and do it again. And again, and again. In fact, we have to do it infinitely many times.9

But, in any case, we’ve split up the ball into a bunch of pieces. For instance, N is associated with all the infinitely many “north poles” we picked, along with the line segment underneath them, and a similar set of line segments for every other word, or set of rotations.

Now we can use the trick from the last post.

Let’s call the set $S(L)$ the set of all the points and line segments in the ball represented by words that start with L, i.e., where the last rotation was to the left. We can similarly define $S(R)$, $S(F)$, and $S(B)$. The set $N$ will just represent all the points associated with the north poles.

The set $S(L)$ is all the points found by rotating the sphere, where the last rotation was to the left. So, if we take all those points, and rotate them to the right to undo that last rotation, as with the graph, we get all the points in $S(L)$, $S(F)$, $S(B)$, and $N$, exactly like we did for the branched graph! As before, we call the left-last points, rotated right, $R\,S(L)$.

We can do a similar thing, and $F\,S(B)$, the points found by rotating backwards last, but then having that last rotation undone, is all the points in $S(B)$, $S(R)$, $S(L)$, and $N$ put together!

That, right there, is the heart of the Banach-Tarski paradox.

It’s easy to get the two balls from what we’ve done. To make the first ball, take all the points in $S(L)$ and $S(R)$, rotate $S(L)$ to the right to get $R\,S(L)$, then put $R\,S(L)$ and $S(R)$ together, and you have a ball! The second ball is similar: take all the points in $S(B)$ and $S(F)$, rotate $S(B)$ forward to get $F\,S(B)$, then put those two sets together, and you have the second ball!

So, there you have it. By cutting up the sphere into a few pieces, and then just rotating and moving them around, you can turn one ball into two!

Admittedly, we’ve glossed over a few important details if you want this to work out perfectly, but I think they can be hidden in a footnote.4

This is quite the paradox! You shouldn’t be able to cut a ball into pieces and put them back into two spheres!

From a physical point of view, this process is, of course, impossible. Not only are the sets we’re cutting the ball into hopelessly complicated and delicate, but they assume that matter is infinitely divisible, which is false. (Subatomic particles are, after all, a particular size, and it’s hard to cut a quark into pieces…)

But even from a mathematical view point, this seems like it shouldn’t work. And, so, if we think that way, we can look back at our assumptions, and see which of the axioms we used seems the most questionable, and try to get rid of that assumption.

What’s fascinating about this proof is that the key problematic axiom is so innocuous that, if you didn’t know what you were looking for, you would probably never find it. The step that is the most questionable is the one where we choose points as new north poles.

The thing is, when we make that choice, there’s no reason to pick one point over another. They’re all just as good as any other. Plus, we have to make infinitely many of these choices, which is also a bit… uncomfortable.

Doing this requires the Axiom of Choice, perhaps the most infamous axiom in mathematics.

And all it says is that you can choose things.

In the next post, we’ll take a look at the axiom of choice and why it’s so important… and infamous. (Assuming the Missus can find the time to draw…)

An excellent video on this paradox, and its proof, can be found on Vsauce’s Youtube channel. In the description, he also lists many resources which I found useful in preparing this post.

Oh, life plans. How fleeting they are.

Since the last post, we’ve had Thanksgiving, a funeral with associated trip, finals, the flu, bad colds, Christmas, the start of a new term, 10 interviews, and so forth. It’s been… busy. That would explain some of it. But another big thing is the complete upheaval of my life plans!

See, to become a professor, after you get a PhD, you usually spend 2 or 3 years at one or two universities as a “postdoc,” which is what I am now. These positions are temporary, and are not expected to lead to permanent positions at said universities.

So, I applied to permanent jobs this school year. Lots of them. I had a bunch of interviews, but I didn’t end up getting hired as a professor.

I could probably scramble and get another postdoc at another university and then do another cycle or two of applications for professor jobs, but… well, academia is stressful. Awesome, to be sure, but stressful too. (When I talked to my mentor about his career path, his frequent use of the phrase “panic mode” didn’t exactly encourage me.)

So, after a lot of thought, I’ve decided to leave academia, and become a computer programmer instead. Of course, since I have limited experience, that means I have till my contract at the university ends to learn enough programming to get hired somewhere. It turns out that covering years of computer science education on my own is time-consuming.

For obvious reasons, then, this blog, though it will probably continue, is going to be updated less frequently in the coming year.

1. The Missus, here: it was my fault. Apparently one cannot blog without cute pictures from wife.
2. Well, okay, in real life, it’s not like you’re going to be able to duplicate balls of gold with this in a get-rich-scheme or anything, but mathematically it works!
3. In case you’ve forgotten, radians are just another unit for measuring angles, like Fahrenheit and Celsius are different units for measuring temperature. By definition, a circle has 360 degrees, or $2\pi$ radians. A triangle’s angles (on a flat surface) add up to 180 degrees, or $\pi$ radians.
4. Well, even here, I won’t go over the details of the proof, but I’ll at least mention some things we glossed over. The most obvious is, perhaps, that we never worried about the very center points of the sphere. You can take care of that with a trick like we did by rotating the circle to fill in a point. Also, you may not have noticed, but we only used $S(L)$, $S(R)$, $S(B)$, and $S(F)$ in making the two copies. We never used the original copy of $N$! So, actually, we ended up with 2 spheres, plus some left over stuff. If you cut up the sphere in a slightly more complicated way (it’s not that bad), you can make sure you don’t end up with this junk. The most subtle problem is that not all points of the ball rotate when we do our rotations. This ends up meaning that we’ve missed some points. But there’s a way to take care of these points as well. More details can be found on the Wikpedia page, or in this short paper
5. This boils down to a variation of the Hilbert’s Hotel paradox, which we talked about way back in our very first post, Infinity plus one
6. Recall that we don’t allow letter combinations that would undo each other, like LR or BF.
7. It takes a bit more work than the circle example to be one hundred percent sure that points don’t overlap after any rotation, but this angle was chosen since it makes the math not too horrible. A proof that it doesn’t overlap can be found on the second and third page of this paper
8. The fundamental problem is that our rotations only let us get countably many points on the surface of the ball, while the surface itself has uncountably many points. As we talked about in How Long is Infinity?, countably many is essentially nothing compared to uncountably many.
9. More specifically, we have to do it uncountably infinite times in order to get them all.

## Double for Nothing: the Banach-Tarski Paradox

Though those stodgy engineers and accountants may disagree with me, one of the most exciting parts of mathematics is learning about a result so bizarre, so against your intuition built up by years of experience, that it must obviously be false. These are often labeled “paradoxes.”

But this is mathematics we’re talking about. These results may assault the intuition, but it’s not because they’re wrong or the result of faulty logic, like so many other paradoxes. We come with proofs and irrefutable logic. Sure, you might be able to argue against the assumptions (axioms) we start with, but given those perfectly reasonable and seemingly innocent rules, we can create truly bizarre things.

No matter how much those accountants dislike it.

And today, we get to talk about a doozy of a paradox. We’re going to take a ball, cut it up into 5 or so pieces, move them apart, rotate them around a bit, and end up with two balls, exactly identical to the original.

This is the infamous Banach-Tarski paradox.

Of course, creating something from nothing is an old trick. For instance, if you take all the numbers between 0 and 1, which has length 1, and you multiply each of those numbers by 2, you end up with all the numbers between 0 and 2, which has length 2! Pretty much everyone’s okay with this one.

But the Banach-Tarski paradox is much weirder. We turn one ball into two, but there is no stretching involved. The only things we do is cut the ball into pieces, move them apart, rotate the pieces a bit, then move them back together. And just moving parts of a ball shouldn’t create a new ball out of thin air.

Right?

But if we do it just right, we can.

So, how do we start this black magic?

While this paradox seems to be about geometry and measuring things, the key step, the key observation, has nothing to do with either. It’s all about a dictionary with every possible word in it.

For simplicity (and to line up nicely with what we’ll do later), let’s pretend our alphabet only has 4 letters — B, F, L, and R. So, the first word in our dictionary would be B, followed by BB, then BBB, then BBBB, then BBBBB, then…

And, finally, after all those exciting words, we’d finally get to BF, then BFB, then BFBB, then…

Of course, after you finish all the B words, you get to start the F words! F, FB, FBB, … FFB, FFBB, … FL, FLB, FLBB, …

I think you get where this is going.

Now, we have all of these infinitely many words, every word that could every be thought of or written. How are we going to print this dictionary?

Well, the first volume of our dictionary could contain all the “B” words. Of course, if everyone knows it’s the “B” volume, we could save space by not printing the first letter of each word. That means the first word would be ___, then B, then BB, then BBB, then …, then F, then FB, then FBB, then …

But wait a second. Those words F, FB, FBB, etc. which represent BF, BFB, BFBB, etc. are actually all the words that begin with F!

In fact, since every word could be extended to a “B” word by just adding a B as the first letter, by taking off that first B, our “B” volume is really a dictionary of all the possible words!1 One volume of our dictionary can take the place of four!

See what we did there?

No? Well, let’s look at it in a more graphical way, and even closer to what we’ll actually do for the ball.

Grab a ball.

Instead of letters in a word, we can think of B, F, L, and R as being directions to rotate the ball a few degrees. “B” means to rotate the ball a bit backwards, towards yourself, “F” means to rotate the ball a bit forwards, away from yourself, “L” a bit left, and “R” a bit right. Then, all the one letter “words” represent rotating the ball exactly once, and we could graphically represent them like this:

The center intersection, which we’ve labeled “N” is not rotating at all. In our dictionary it was the ____ “word.”

Two letter “words” would represent rotating twice. For words, though, we could have something like “BF” or “FB,” but when thinking about rotations, we don’t want to undo the rotation we just did, so we’re not going to allow rotations that undo the one we just had. In fact, we can think of them cancelling out, so that “BF” is really the same as “N.”

But the remaining ones, like “FL” and “RF” are fine. We can add those to our graph like this:

We make each additional step half the size so that we can keep all these words straight, and when we say “FL,” we should think of that word representing doing a left rotation, then a forward rotation. Thus, all the series of rotations that end with a left rotation are on the left hand side of our graph.

After the two letter words, we get the 3 letter words (3 rotations), then the 4 letter words (4 rotations), etc.. This graph contains all possible order of rotations we could have made! Each intersection on the graph represents a different order of rotations.

Cool!

So, how can we create something out of nothing?

Look closely at, say, the left part of the graph.

All of the points in this part of the graph are represented by words starting with L, meaning the last rotation was to the left. But, like with the dictionary, by removing the first letter (the last rotation), we can have other words appear!

For words, it made sense to just remove the first letter. But since now we’re doing series of rotations, we can remove the first L instead by taking the ball and rotating right to undo that last left rotation!

If we do that, for instance, L becomes RL, which cancel out and give us the center point, N. The rotation LF becomes RLF, but the R and L cancel out and this becomes just F. The rotation LBR becomes RLBR, which is really just BR.

If you keep track of each set of rotations, after undoing that last L, the set of rotations on the left (which was about 1/4 of the graph) becomes the entire top, left, and bottom parts of the graph (about 3/4 of the graph!)

The only points we don’t get are the ones on the right, but that’s because those words start with R, and we couldn’t have had “LR…” words in the left part of the graph, since we didn’t allow rotations that would cancel each other out like that.

So, by undoing a rotation, we seemingly create points out of thin air! This is the key trick that will let us duplicate a ball.

But for right now, let’s use it just on this graph. I’m going to cut the graph up into 5 pieces, then, using this “undoing a rotation” trick, make two complete copies of the graph, with one piece left over!

To do this, we’ll call $S(L)$ to be the set of all rotations, where the last rotation was to the left. This is the set we were playing with earlier. We’ll define $S(R)$, $S(B)$, and $S(F)$ similarly. The fifth and final set is the odd one out, just the center point N.

To undo the final left rotation of $S(L)$, we can rotate to the right, which we could write as $R\,S(L)$. As we’ve already said, $R\,S(L)$ is already the entire graph, except for the right part, $S(R)$. But that was one of the pieces we have lying around. So, the first copy of the graph is $R\,S(L)$ plus $S(R)$.

To make the second copy, we can take all the rotations that end with a back rotation, $S(B)$, and then undo it with a forward rotation. Then, like before, $F\,S(B)$ plus $S(F)$ make a second copy of the graph.

Two graphs for the price of one!

This trick is the true heart of the Banach-Tarski paradox. By using rotations to split up the ball in a very careful way, we can create points out of nothing by undoing the last rotation. And then we can very carefully put them back together, creating two balls out of one!

But let’s leave those details till next time!

1. Plus that extra ____ word.

## Non-measurable Sets That Go Bump in the Night

It’s Halloween! Which means that we here at Infinity Plus One get to do something spooky!

That’s right, we’re going to talk about sets of numbers so weird that even the very idea of length breaks when looking at them.

In the last few posts, we’ve been talking about how to measure the lengths of sets, even ones that are weird.

In How Long is Infinity?, we introduced how we measure things — by covering up a set with little intervals, and then calling the length of our set the smallest lengths of intervals that cover our set.

Using this length, it turns out that any countable set, like $\{1, 2, 3, \cdots \}$, or even the set of all rational numbers, has zero length.

In The Cantor Set, we showed that, though uncountable sets like [0,1] (all the numbers between 0 and 1) have positive length, there are uncountable sets with zero length. The Cantor set is the main example.

In this post, we want to look at non-measureable sets. Sets which are so weird that they break our “ruler” and make it impossible to make any sense at all of their length.

As a technical caveat, the “ruler” we’ve been discussing so far is technically the “outer Lebesgue measure,” which is not really the same as the “Lebesgue measure” that mathematicians use. However, the difference is buried in technical details that would distract from the story, so we’ll bury those important details for this post.

So what does an non-measureable set look like?

It’s gotta be weird. The definition of length, or measure, that we have is pretty robust. It can handle some pretty weird sets, like all the decimals with a 7 in them, and spit out a length.1

So, in order to come up with a non-measureable set, we’re going to have to work hard.

What we’re going to do is come up with a weird set, and we’ll prove that if we add up infinitely many copies of it, somehow that total length will end up between 1 and 3. But that can’t be right, since adding up infinitely many of the same number always gives either 0 or infinity!2

To start, we’re going to split all the real numbers into groups.

The first group is all the rational numbers, i.e., any number that can be written as a fraction of integers, like 2/3 or -712/2341. We’ll call this set $\mathbb{Q}$, for “quotient.”

The other groups are all copies of the rational numbers, but shifted left or right by a different real number. For instance, we could have the group $\pi + \mathbb{Q}$, which is the set of all numbers which are $\pi$ plus any rational number you want. Or you could have $e + \mathbb{Q}$, which is the set of all numbers which are $e$ plus any rational number you want.

There are a whole lot of these groups, and every real number is in one of them. On the other hand, there is more than one way to name which set you’re talking about.

When we gave the two examples of these groups, we used $\pi + \mathbb{Q}$ and $e+\mathbb{Q}$ to define them. In other words, we picked a particular number, $\pi$ or $e$, that happened to be in the group, and used it as a representative of that set.

But there’s more than one number in $\pi+\mathbb{Q}$, and we could have used any of them to represent the set, not just $\pi$. For instance, $(\pi-1/3)+\mathbb{Q}$ is the exact same set, with the exact same numbers in it. So is $(\pi+712/2341) + \mathbb{Q}$ or $(\pi+17) + \mathbb{Q}$

…though $\pi/2+\mathbb{Q}$ would not be the same, since $\pi$ and $\pi/2$ do not differ by a rational number.

As long as our representatives differ from each other by a rational number, the sets are exactly the same.

Using these groups of numbers, we can now construct the unmeasureable set.

Let $V$ be the unmeasurable set. To construct it, look at each of the groups of numbers we came up with earlier. From each one, pick a single representative that happens to be between 0 and 1. For instance, from the set $\pi+\mathbb{Q}$, we could pick the representative $\pi-3$, which is between 0 and 1. From the set $\mathbb{Q}$, we could pick 0 or 1 or 1/2 or any rational number between 0 and 1.

Now that we’ve chosen one representative from each group, it turns out that $V$ is unmeasurable!

Here’s how we’ll prove it.

Similar to how we took $\mathbb{Q}$ and moved all the numbers by $\pi$ to make $\pi+\mathbb{Q}$, we can take our set $V$ and move all the numbers by a rational number $q$, and make a new set $q+V$.

To make things clearer, this means that if a number $x$ is in $V$, then the number $q+x$ is in $q+V$. And, in the opposite direction, if a number $y$ is in $q+V$, that means that $y-q$ must have been in $V$.

But there’s something funny about $q+V$. No matter how small $q$ is, $V$ and $q+V$ never overlap!

Remember, each of the groups we came up with earlier had infinitely many different representatives we could have picked. But the representatives had to differ from each other by a rational number if they were supposed to represent the same group.

If $V$ and $q+V$ overlap, that means there would be a number $x$ in $V$ and $q+V$. That means that $x-q$ would also have to be a number in $V$. Thus there are two numbers ($x$ and $x-q$) that are both in $V$, but differ by a rational number.

But remember that the numbers in $V$ are representatives of our groups, and so if they differ by a rational number, they represented the same group.

But we only picked one representative from each group to put in $V$.

And so $V$ and $q+V$ can’t overlap!

Next step: Put the rational numbers between -1 and 1 into some order. There are infinitely many of them, but they’re still a countable set, so we can do it. There are more details in the earlier post The size of infinity, but here’s the kind of ordering we could use to make sure we get them all.

Since we have an ordering for the rational numbers between -1 and 1, we’ll call $q_1$ the “1st” rational number, $q_2$ the “2nd,” etc., and $q_k$ the “k-th” rational number. Then, we can come up with a whole bunch of copies of $V$ moved around. We’ll call $V_k$ the the set $q_k +V$, i.e., the set $V$ moved up or down by $q_k$.

Just as before, none of the $V_k$ overlap. Also, since $V$ only had numbers between 0 and 1, and $q_k$ is between -1 and 1, then all the numbers in $V_k$ are between -1 and 2.

The more difficult part is to recognize that if you put all of the $V_k$ together (“take their union”), then together, they contain every number between 0 and 1.

To see this, pick any number $x$ between 0 and 1. No matter which number we pick, it’s in one of the groups we made earlier, perhaps $\pi + \mathbb{Q}$. But, when we made $V$, we picked one representative $r$ (between 0 and 1) for each of these groups. Since the representative $r$ and the number $x$ are in the same group, they have to differ by a rational number, and since they are both between 0 and 1, that rational number they differ by has to be between -1 and 1. That means that $x$ is in the set $V_k$ that happens to be $V$ moved by $q_k = x-r$, which is a rational number!

Yeah, it’s kind of hard to keep all these sets straight, but we’re almost done.

To finally see that the set $V$ can’t be measured (i.e., is non-measurable), let’s pretend that we can measure it, and show that something impossible happens.

If we can measure $V$, the sets $V_k$ have the same length, since they’re really the same set, just moved up or down on the number line.

Since, put together, the $V_k$ contain all the numbers between 0 and 1, their total length has to be at least 1. And, since the $V_k$ only have numbers between -1 and 2, clearly their total length has to be no bigger than 3. If we wrote $L(V_k)$ to represent the length of $V_k$, we could write that like this:

$1 \leq L(V_1) + L(V_2) + L(V_3) + \cdots \leq 3$

But, again, the set $V$ has the same length as each of the $V_k$, and so, really, we’re saying:

$1 \leq L(V) + L(V) + L(V) + \cdots \leq 3$

But we’re adding up infinitely many of the same number! If the length $L(V)$ were 0, adding up infinitely many zeros gives zero length. If the length $L(V)$ were any number bigger than zero, adding up infinitely many of them would give infinite length!3

And so, since 0 and $\infty$ are not between 1 and 34, we have shown something impossible. Thus $V$ cannot be measurable. We have broken our ruler.

So, yeah, non-measurable sets are weird. And we had to do a lot of work to come up with one.

But, in the end, it might seem like a waste of effort. I mean, it’s just a weird set that no one in their right mind would care about anyway.5

But there are some weird things you can do with non-measurable sets.

The most famous is the Banach-Tarski paradox. There is a way you can take a sphere, cut it up into a few pieces, and rearrange them, and end up with two spheres, exactly identical to the original.

But that’s for next time!

Happy Halloween!

1. For all decimals between 0 and 1 that have a 7 in them, the length it gives is 1. In other words, virtually “every” number has a seven in it!
2. The name for the set we’ll be constructing is a “Vitali set.” You can read more on the Wikipedia page
3. Importantly, there are only countably many sets and lengths that we’re considering, so there’s no funny business going on. Adding up countably many numbers or sets works the way you think it “should.”
4. Citation needed.
5. Of course, mathematicians are not really known for ever being in their right mind…