No interval in it, and length to spare

The middle-thirds construction removes a third, then two ninths, then four twenty-sevenths, and the pieces removed add to exactly one. That is why what is left has no length, and it is also why the argument is a little too convenient.

Nothing about the construction requires the removed middles to be thirds. Take a quarter first, then an eighth of each piece, then a sixteenth, and the same argument produces a set with the same shape and a completely different size.

The arithmetic of the fat construction #

At stage $k$ there are $2^{k-1}$ intervals, and a fraction $1/(4 \cdot 2^{k-1})$ of each is removed. The length taken at that stage is therefore the number of intervals times each one’s length times that fraction, which works out to $1/8$ at the first stage after the initial quarter, and the total removed converges.

The clean way to state it is multiplicatively. What survives stage $k$ is a fraction $1 - 1/(4 \cdot 2^{k-1})$ of what was there, so the length remaining after all stages is the infinite product

$$\prod_{k=1}^{\infty}\left(1 - \frac{1}{4 \cdot 2^{k-1}}\right),$$

and the figure computes it to sixty terms and finds it is about $0.5776$ — bounded away from zero rather than converging to it.

A product of terms below one need not converge to zero, and that single observation is the whole difference between the two constructions. For middle thirds every factor is $2/3$ and the product is $(2/3)^k \to 0$. Here the factors approach one fast enough that the product does not.

Why an infinite product can stay positive #

The criterion is worth having, because it is exactly what separates the two cases.

Taking logarithms turns the product into a sum: $\log \prod (1 - a_k) = \sum \log(1 - a_k)$, and for small $a_k$ that is about $-\sum a_k$. So the product is positive exactly when $\sum a_k$ converges.

For middle thirds, $a_k = 1/3$ every time and the sum diverges, so the product is zero. For the fat construction $a_k = 1/(4 \cdot 2^{k-1})$, a geometric series with sum $1/2$, so the product is positive. A series whose terms vanish may still diverge, and the whole question here is whether this particular one does.

That is a satisfying place for the distinction to live: not in any geometric feature of the sets, which are indistinguishable, but in whether one series converges.

The set still contains no interval #

The second half of the claim is that nothing has been gained in the way of intervals, and it is checked by watching the longest bar.

At each stage every surviving interval is cut, so the longest one shrinks. The figure asserts that it strictly decreases at every stage and reports its value at the last drawn one, which is already under a hundredth. In the limit the longest surviving interval has length zero, so no interval of positive length survives.

A set containing no interval is called nowhere dense — more precisely, its closure has empty interior, and this set is closed, so the two amount to the same. So the fat Cantor set is nowhere dense and has measure $0.5776$, while the thin one is nowhere dense and has measure zero.

Two properties that always travelled together have been separated by one example, and that is the entire reason to build it.

Reading the numbers off the picture #

The hero prints a length beside every row, and it is worth following them down because the two columns behave completely differently.

The fat construction runs $1$, then $0.75$, then $0.6563$, $0.6152$, $0.596$, $0.5867$, $0.5821$ — falling by less each time and visibly heading somewhere positive. The middle-thirds column beneath runs $1, 0.6667, 0.4444, 0.2963, 0.1975, 0.1317, 0.0878$, falling by a constant factor each time and heading to nothing.

The distinction is between a sequence whose successive ratios approach one and a sequence whose ratios are constant. Both are decreasing and both are bounded below; only the second has zero as its limit.

Reading a table of decreasing numbers and deciding where they are going is exactly the skill this rung is about, and the two columns are the same table with different answers. Six rows are not enough to be certain of either, which is why the generator computes sixty terms of the product and asserts the result rather than leaving a reader to extrapolate from the drawing.

What the two sets have in common #

It is worth listing the properties the two sets share, because the list is long enough to be surprising.

Both are closed. Both are bounded. Both contain no interval. Both have no isolated points — every point of either has other points of the set arbitrarily close. Both are uncountable, with exactly as many points as the whole line. Both are homeomorphic to each other and to the space of infinite binary sequences: there is a continuous invertible correspondence between them with continuous inverse.

That last one is the strongest. As topological spaces the two sets are literally the same space, and every property expressible in terms of open sets alone must therefore agree on them. Measure is not such a property, and this pair is the standard proof that it is not.

The consequence for anybody carrying intuition between topology and measure is direct: small in the sense of containing no interval and small in the sense of having no length are independent notions, and an argument that slides between them is broken.

Where each notion of small comes from #

The two notions answer different questions, and the questions are worth separating.

Topological smallness asks whether a set can be avoided — whether every interval contains points outside it. That is a question about position, and it is preserved by any continuous deformation with a continuous inverse.

Measure asks how much of the line a set takes up when the line is measured with a ruler. That is a question about size, and it is destroyed by deformation: stretching the middle-thirds set by the right continuous map turns it into a set of positive measure, which is exactly what the correspondence between the two sets does.

Neither notion is more fundamental. The dimension of the middle-thirds set is a third notion again, and gives yet another answer — and the fat set’s dimension is one, since it has positive length, so even that does not separate the two constructions in the way one might expect.

How fat can it be made #

Once the parameter is loose the natural question is how much can be kept, and the answer is everything short of all of it.

Removing a middle of proportion $1/(m \cdot 2^{k-1})$ at stage $k$ leaves a product whose value climbs towards one as $m$ grows: at $m = 4$ it is $0.5776$, at $m = 10$ about $0.81$, at $m = 100$ about $0.98$. The set is nowhere dense at every $m$, because the longest bar still shrinks to nothing, so a nowhere dense subset of the unit interval can have measure as close to one as anybody likes.

What it cannot have is measure exactly one. A closed set of full measure in the interval and with empty interior would have a complement that is open, dense and of measure zero — and an open dense set is a union of intervals covering everything, whose total length cannot be nothing.

So the supremum is one and it is not attained, which is a shape worth recognising: a family of examples approaching a bound that the bound’s own definition forbids. The same shape appears whenever a construction is pushed to its limit and the limit object fails to exist.

The consequence is a slightly alarming picture of the unit interval: it can be split into a nowhere dense closed set holding $99\%$ of the length and a dense open set holding the remaining $1\%$. Density and length have nothing to do with each other.

The complement, which is the more familiar object #

Turning the picture round makes the fat Cantor set look less exotic.

Its complement in the unit interval is an open set: a countable union of the removed middles, dense in the interval, and of total length about $0.4224$. So the fat Cantor set is the complement of a dense open set of measure less than one.

Dense open sets of small measure are easy to build directly. Enumerate the rationals, put an interval of length $\varepsilon/2^n$ round the $n$-th, and the union is dense, open, and of measure under $\varepsilon$. Its complement is closed, nowhere dense, and of measure over $1 - \varepsilon$ — a fat Cantor set by another route.

That construction is where the next rung starts, and it is worth noticing that it produces the same phenomenon from the opposite direction: the middles here are chosen to shrink, and there they are chosen to cover a countable set cheaply.

The addresses still work, and say nothing about size #

The middle-thirds set has a famous second description: it is the numbers whose base-three expansion can be written without a $1$. That description is what makes its uncountability easy, and it is worth asking what happens to it here.

The fat set has an address system too, and it is the same one: at each stage a point is in the left piece or the right piece, so a point is named by an infinite sequence of lefts and rights, and different sequences name different points. That is what the homeomorphism with the binary sequences amounts to.

What changes is the arithmetic of the addresses. For middle thirds the address is a base-three expansion and the correspondence is exact, which is why the set can be paired off with the whole interval by a rule about digits. For the fat set the pieces at stage $k$ have lengths that are not powers of anything, so the address is combinatorial and not arithmetical.

The uncountability survives and the digit arithmetic does not, and the length of the set is invisible in the addresses either way — the two sets have identical address structures and different measures, which is the same fact as their being homeomorphic.

Smith, Volterra, Cantor #

The set is usually named after Cantor and he was third to it.

Henry Smith published the construction in 1874, in a paper on the integrability of discontinuous functions, and used it for exactly the purpose the next rungs will: to show that a bounded function can be discontinuous on a set of positive measure and so fail to be Riemann integrable. Volterra used it in 1881 to build a differentiable function whose derivative is bounded and not Riemann integrable — which is a serious defect in the Riemann integral and was one of the reasons a better one was wanted.

Cantor’s own construction, published in 1883, was the middle-thirds one, and it was aimed at a different question: he wanted an uncountable set that was nowhere dense, to show those two properties were compatible.

Three people building nearly the same object for three unrelated reasons is a good sign that the object is important. It also explains why the fat version is the less famous of the two despite being the more useful counterexample: it was found first, for a technical purpose, and named after somebody else.

Volterra’s function, which is what it was for #

Volterra’s construction deserves a sketch, because it is the sharpest demonstration that the Riemann integral is inadequate.

Take a fat Cantor set. On each removed interval, build a small bump that is differentiable, that oscillates near the ends of the interval, and whose derivative oscillates between $-1$ and $1$ without settling. Do that on every removed interval, and define the function to be zero on the fat Cantor set itself.

The result is differentiable everywhere, with a bounded derivative. And the derivative is discontinuous at every point of the fat Cantor set — a set of positive measure — so by the criterion the fourth rung is about it is not Riemann integrable.

So there is a function whose derivative exists everywhere, is bounded, and cannot be integrated by the theory that was supposed to invert differentiation. The fundamental theorem of calculus fails, not because of some pathology at infinity but because of a set of positive measure that contains no interval.

Fat Cantor sets inside every interval #

One more property, because it is what makes the construction useful rather than merely surprising.

The construction can be run inside any interval, at any scale, and it can be run inside the removed middles of a previous run. Doing that repeatedly builds a set that is nowhere dense in a stronger sense while still holding most of the length, and it is how counterexamples in this subject are usually assembled: choose the scales so that the total length behaves as required, and the topology takes care of itself.

That flexibility is why the set turns up wherever a counterexample is needed. A function that is continuous exactly off a fat Cantor set, a derivative that is bounded and not integrable, a set whose boundary has positive measure — each of these is a case where the intuition trained on intervals fails, and each is built by putting something on the removed middles and nothing on the set itself.

The generic move is: build the set to control the measure, then define the function on its complement. The complement is a countable union of intervals, and a function defined separately on each of countably many intervals is as easy to control as one function on one interval.

What the pictures cannot show #

Every drawing stops at a finite stage. At stage six the sixty-four bars have positive width, and the set is the intersection of all the stages, with no width anywhere. There is no stage at which the difference becomes visible, and there never will be.

The two sets look identical and are not. That is the whole content of the hero figure, and it is the one thing a picture genuinely cannot convey: two rows of bars, arranged the same way, whose totals differ by a factor of six and whose limits differ by everything.

And the infinite product is computed rather than drawn. Sixty terms of it, in the generator, well past the six stages the picture holds. A reader looking at six rows is looking at evidence that the total is coming down slowly, not at the number it is coming down to.

Where the ladder goes next #

The construction above assumed that lengths could be added up — that removing countably many intervals from an interval leaves something whose length is the difference. That is not a definition, and the next rung supplies one: outer measure covers a set from outside and takes the smallest total length that manages it, which is the definition every statement on this page has been using informally.

Sideways: the middle-thirds set is where this construction started, and the geometric series is the arithmetic that decides which of the two happens.

What is worth carrying away #

When two properties always travel together, the useful move is to build an object with one and not the other.

Contains no interval and has no length agree on every set most people meet, and the agreement is a coincidence of the examples rather than a theorem. Separating them takes one construction and one convergent series, and once separated they can never be confused again.

The construction is also a warning about the middle-thirds set specifically. It is the standard example of a small set, and half of what it demonstrates is an accident of the number three. Anyone reasoning from it should check which half is being used.