# Hopefully Interesting

## December 28, 2007

### Length and Area I

Filed under: math.CA — Pietro @ 6:54 am
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This is a topic which has fascinated me for years, and on which I’ve spent many (fortunately unsuccessful) hours trying to prove theorems that are actually false. I find it to be a good introduction to the weirdness of analysis: you know, the gradual realization that much of our intuition about lines, surfaces and continuity is simply not applicable to the analytical formalization of lines, surfaces and continuity. Some mathematicians, like Weyl and Brouwer, blamed this weirdness on the fact that the intuitive notion of a continuum (the line) is not very well captured by something as disconnected as a set of points (the real numbers).

In any case, we have to develop some new intuitions if we’re going to expect, rather than be surprised by, the Koch snowflake, nonmeasurable sets, and families of curves that have positive area but only at the endpoints.

The story begins with the naïve picture of a closed plane curve (a loop) we had in our minds before we learned analysis. Most of us thought of it as something similar to a circle, with a dent or two thrown in for generality perhaps. Our “general” plane curve looked a bit like a kidney, the body of a guitar, or a mix of both:

A few intuitive facts suggest themselves upon inspection of this picture:

1. The curve itself, the boundary of the kidney, has no area; more precisely, it has zero plane measure.
2. However, it seems to have a length. We imagine the curve as a piece of string we can lift up, straighten out, and measure against a ruler.
3. If the curve doesn’t cross itself (like a figure eight does), then it separates the plane into two parts, the “interior” and “exterior” of the curve.

Are these observations true?

The one correct answer to every single question in the world, including this one, is: it depends on what we mean by the terms employed. Interestingly, different meanings of the phrase “closed plane curve”, all of them reasonable, give different answers to each of the three questions above.

One standard meaning is “the image of a continuously differentiable map from the unit circle into the plane”. Here, “circle” refers to the outer edge of the familiar figure, that is, the round outline that the pencil traces out. The interior of the figure I call a “disk”. (Thus, a circle is to a disk as a ring is to a frisbee.) I prefer to have the domain be a circle instead of the more traditional closed unit interval $\left[ 0,1 \right]$ in order to avoid repeating $\phi(0) = \phi(1)$ every time I want the curve to be closed (which will be pretty much always). Also, special issues of differentiability at the endpoints don’t arise: the whole description is more symmetrical and elegant.

If we use this strong meaning of closed plane curve, 1–3 are answered in the affirmative, proving which is a nifty calculus exercise (especially #3). Even #2, which logically demands a definition of length before it can be approached, yields to the most natural definition that I know of, and which will be the topic of a future post. Not wishing to spoil the reader’s fun, we concentrate on somewhat more general meanings of “closed plane curve”.

Definition. A closed plane curve is the image of a continuous map from the unit circle into the plane, which we will have occasion to write as $\phi : S^1 \rightarrow \mathbb{R}^2$. A simple closed plane curve is one for which the map is injective: these are curves which don’t cross themselves. It’s usual to denote them by $\phi : S^1 \hookrightarrow \mathbb{R}^2$. (Notice how the arrow is different.) Since we shall be working exclusively in $\mathbb{R}^2$, I’ll drop the adjective “plane” from now on.

We might hope that these curves possess properties 1–3. After all, what’s a continuous curve? It’s just a differentiable curve with a few creases, right? How could it not have a length, and worse still, how could it have an area?

Well, it turns out that continuous functions can do some pretty unexpected things. Without differentiability, the map $\phi$ can stretch arbitrarily small arcs of the circle into proportionally large segments of the curve, making its total length infinite, while still lying in a bounded region of the plane. (A technical way to put this is that, without differentiability, there need be no Lipschitz condition.) The Koch snowflake is an example of this situation, and the reader may find some enjoyment in proving that the snowflake really is a closed plane curve, according to our definition.

Even more amazing is that the circle can be ingeniously stretched to cover an entire square. Not just the four edges, mind you, but the whole thing. If you have not heard about this, I recommend reading about space-filling curves, preferably in Hans Sagan’s wonderful book. (The price is a bit steep for this smallish book; I read it at the library.)

Since this topic is extensively discussed in many websites and an entire textbook, I shall not take it up in the traditional manner; I will simply assume that you have a passing acquaintance with it, and try out some variations. Nor will I elaborate on the fact that the third property — that of separating the plane into two regions — is the only one valid in the general setting of continuous $\phi : S^1 \rightarrow \mathbb{R}^2$. That proof alone could fill a post thrice this size.

Alright. Suppose we are baffled by the phenomenon of space-filling curves, and would like to put our finger on where exactly our intuition went wrong. How is it that something so one-dimensional as the circle can be stretched and deformed into something two-dimensional?

One natural candidate for suspicion may be the map $\phi$, described in Wikipedia, that does the trick. After all, the image of the circle is only as one-dimensional as $\phi$ makes it. Indeed, we know that an arbitrary map can turn a circle (or line) into almost anything else, from a plane square to a nine-dimensional ball, since all these sets of points have the cardinality of $\mathbb{R}$, and are thus in bijective correspondence with each other. Therefore, if we are to be legitimately surprised, we must inquire into the nature of $\phi$: in what sense does it preserve the one-dimensional character of a circle?

Since dimensionality of curves and plane regions sounds like a topological concept, we may start by examining how faithfully $\phi$ preserves the topology of $S^1$. Continuity is not delicate enough, since continuous maps may turn an arbitrary object into a single point. It’s a bit surprising that one can “build” stuff with continuous maps, instead of just “collapsing” it, but we’d be flabbergasted if the space-filling map $\phi$ turned out to be that most faithful of topological morphisms: a homeomorphism. Our intuitive notion of dimension would be all but chucked out the window.

Fortunately, that isn’t the case. It’s easy to see that $\phi$ isn’t a homeomorphism, simply because it’s quite strongly non-injective. What’s more, it’s impossible for a homeomorphism to turn $S^1$ into a plane square. This is also kind of unexpected: we can’t map the “thin” space $S^1$ continuously onto the “fat” space $\left[ 0,1 \right]^2$ without repeating values — which, intuitively, only “spends” our already very “scarce” domain set.

What happens is that, for a curve to fill up a square without repeating values (ie, without intersecting itself), it would have to dodge itself in ingenious ways. One can imagine that, after filling up half the square, the curve may sort of paint itself into a corner, and after filling up 99%, this will almost surely happen. Theorem 2, below, shows that such a situation would actually occur as soon as the curve filled up any tiny disk, no matter how small. In preparation, we need the simple

Lemma 1. If $X,Y$ are topological spaces, with $X$ compact and $Y$ Hausdorff, and $f : X\rightarrow Y$ is a continuous bijection, then $f$ is a homeomorphism.

Proof. We show that $f$ is a closed map. Being bijective, it will be a homeomorphism. Indeed, if $K$ is a closed subset of $X$, then $K$ is compact. Therefore, $f(K)$ is compact. Since $Y$ is Hausdorff, $f(K)$ is closed in Y. $\Box$

Theorem 2. Consider a continuous injective map $\phi : S^1 \hookrightarrow \mathbb{R}^2$. The image curve, $\phi(S^1)$, has empty interior.

First proof. Since $S^1$ is compact and $\mathbb{R}^2$ is Hausdorff, lemma 1 applies: $\phi$ is an immersion, ie a homeomorphism between $S^1$ and $\phi(S^1)$ in the induced topology.

Suppose $\phi(S^1)$ contains an open disk $D_p$, centered at $p$. Let $C$ be a circle inside $D_p$.

Since $C$ is closed and connected, its inverse image under $\phi$ is either a point, a closed arc, or all of $S^1$; since $C$ is uncountable, it can’t be a point. This means that, as $\phi$ “traces out” its curve, it can never leave a circle “unfinished”; ie, if $\phi$ touches $C$, it must trace out the whole of $C$ before moving on. We then have two easy contradictions.

First, by tracing out $C$, $\phi$ comes back to the point where it started the tracing. If the inverse image of $S$ is an arc $\widehat{ab}$ in $S^1$, then $a$ and $b$ map to the same starting/ending point in $C$, which hurts injectivity. On the other hand, if the inverse image is all of $S^1$, then the image curve is simply $C$ and has empty interior. Either way, we’re done.

Alternatively, let $S_1, S_2, S_3$ be concentric circles inside $D_p$, each contained in the next. Suppose we start drawing $\phi$‘s curve from $S_2$, and, after tracing out $S_2$, touch $S_1$ before touching $S_3$. Then we’re on the “inside” of $S_2$ but still missing $S_3$ from the image. However, by the Jordan curve theorem, we can’t go outside to finish the job. (This proof ending is circular, because the Jordan curve theorem already contains the statement of our theorem. Still, it gives some intuition.) $\Box$

Second proof. Suppose $\phi(S^1)$ contains the open disk $D_p(\epsilon)$, centered at $p$, of radius $\epsilon > 0$. Let $C_p(\delta)$ be the circle of center $p$ and radius $\delta > 0$. Clearly, $D_p(\epsilon)$ is the disjoint union of all $C_p(\delta)$ with $0 \leq \delta < \epsilon$. Since the $S_p(\delta)$ are connected, their inverse images in $S^1$ must be single points or arcs. The ones which are not points contain open arcs.

Now, there are uncountably many $S_p(\delta)$, and only one of them consists of a single point ($\delta=0$). Therefore, there are uncountably many pairwise disjoint open sets in $S^1$. But $S^1$ is second-countable, a contradiction. $\Box$

Third proof. Suppose $\phi(S^1)$ contains an open disk. Then one can remove two points from it (and thus from $\phi(S^1)$) while still keeping it path-connected. This is impossible for $S^1$. $\Box$

An easy consequence of theorem 2 is that space-filling curves can never be homeomorphic to a circle, which assuages our suspicions that the intuitive notion of dimension might be topologically inadequate. (In fact, there are purely topological notions of dimension of a space.) We may even conjecture that any plane set homeomorphic to a circle has zero area. This is a bold conjecture, since topology and measure theory often don’t mix all that well, and it doesn’t follow from theorem 2 because a plane set may have empty interior but still have nonzero area, e.g. the irrational points of $\mathbb{R}^2$.

Lemma 1 suggests where to continue our search. If we want a bona fide one-dimensional object with positive area, we should look at the simple closed plane curves: injectivity is enough to guarantee that a continuous map will be a homeomorphism, and thus faithfully preserve dimensionality. So an initial idea would be to fiddle with a space-filling curve and try to remove its self-intersections, in a way that leaves it still occupying “most” of the unit square.

Let’s look at the first few iterations of one of the simpler such curves, to see what kind of self-intersections it has:

The iteration mechanism should be clear: at the $n$-th step, we picture the unit square as being divided into $4^n$ tiny subsquares, each of which contains a triangle, as in the leftmost figure (possibly turned on its side, or upside-down). We then replace the triangle with a (scaled and possibly rotated) copy of the middle figure. Note that the bottom of the outer square is also part of the curve: we’re mapping $S^1$ into the unit square.

As most other space-filling curve constructions, ours divides the unit square into subregions at each step, and subsequent iterations only refine the curve inside individual subregions. This is what allows us to prove existence of a limit curve through completeness of the space $\mathcal{C}(S^1,[0,1]^2)$, of continuous functions from the unit circle into the square, with the uniform metric: increasingly local changes to a curve yield a Cauchy sequence of curves.

For example, in the middle figure above, we have an approximation to the final curve, which will change in the next steps. However, the first quarter of the curve, corresponding to the bottom-left triangle and half the base of the outer square, will always remain within the bottom-left subsquare.

This gives us a clue: intersections between different quarters, sixteenths, … of the curve can only occur on the boundaries of the quarter, sixteenth, … subsquares! So what if we “push” the next iteration away from the boundary, at each step, forming “windows” like so?

This procedure removes self-intersections at each iteration, and a simple argument shows that the limit curve is also free of self-intersections. The only problem is that, unless we choose the “pushing” mechanism carefully, the limit curve may end up having zero area. For example, in the pictures above, I made the “pushing” as simple as I could: instead of splitting each square into four pieces half the size, I split them into four pieces two-fifths of the size, centralized. This means that the sum of the areas of the subsquares decreases by a factor of $4 \times (2/5) ^2 = 16/25$ after each step. Thus the limit curve, which lies in the intersection of all the subsquares — except for the denumerably many “connecting segments”, each of which has zero area — has zero area itself.

What we need is a way of “pushing” which becomes very small, very quickly, so that the intersection of all the subsquares still has positive area. Since we are very free in this respect — any amount of pushing at all will remove self-intersections — we may just do it. For instance, at the $n$-th step, we split each subsquare into four pieces $1 - 1/(n+1)^2$ times the size. It is easily established that, if a sequence $(a_n)$ of non-negative reals has sum less than $1$, then the infinite product of the sequence $(1-a_n)$ converges and

$\prod_{n\in\mathbb{N}} (1 - a_n) \geq 1 - \sum_{n\in\mathbb{N}} a_n > 0$

Therefore, the latter method of pushing yields a limit curve that occupies area at least $0.35$. (Technically, we would have to show that every point in the intersection of all the subsquares is also a point in the limit curve; but this follows by the standard arguments that are used to prove the space-fillingness of the usual curves, e.g. Peano’s and Hilbert’s.)

When it is first suggested, the mind boggles to think of the Jordan curve theorem applied to a curve of positive area. However, the pictures above give a nice intuition of what goes on: any point which is not on the curve gets left behind on some “windowsill”, and on those the theorem is rather understandable.

So there we have it: a homeomorphism of the unit circle onto a subset of the plane of positive area. In a follow-up post we will investigate possible definitions of the length of a curve, and show that, fortunately, a curve of finite length must have null plane measure.