Sierpinski's Triangle, Holes Inside Holes, math that surprises you lesson

Take a filled equilateral triangle, join the midpoints of its sides and cut out the middle triangle. Then do the same in each of the three triangles that are left, in each of the nine after that, and so on without end. The number of pieces triples at every stage, so does much of the triangle survive, or almost nothing? This page counts the pieces and the area at every stage, shows that the area left over shrinks to zero, and explains why the set that remains still contains the whole outline of the triangle and holds together in one piece.

Key terms

Sierpinski triangle

Definition The Sierpinski triangle is the set of points that survive the following construction. Stage 0, written $S_0$ , is a filled equilateral triangle with its edges included, and its area is $A_0$ . To pass from one stage to the next, take every filled triangle of the current stage and join the midpoints of its three sides. This cuts it into four triangles with half its side length, three at its corners and one in the middle that points the other way. Remove the inside of the middle triangle but keep its three edges, and keep the three corner triangles with their edges. Removed triangles are never cut again. Stage $n$ , written $S_n$ , then consists of $3^n$ filled triangles whose side is $2^{-n}$ times the original side, and the Sierpinski triangle $S$ is the set of points that belong to every stage. The surprise is in what the limit keeps and what it loses. Every stage has the positive area $A_0 (3/4)^n$ , but $S$ has zero area. Yet $S$ is far from empty. It contains the three sides of the original triangle and the edges of every triangle of every stage, and it is connected, since any two of its points can be joined by a path that stays inside $S$ . The Polish mathematician Wacław Sierpiński described the set in 1915.

In plain words Each cut keeps three of four equal pieces, so a quarter of what is left disappears at every stage, and after many stages almost no area remains. The cuts only take away the insides of triangles, never their edges, so the outline and a whole network of inner edges are never removed.

Example The table follows the first stages, with sides and areas given as fractions of the original side and area.

Stage	Triangles	Side of each	Area of each	Total area
0	1	1	1	1
1	3	$1/2$	$1/4$	$3/4$
2	9	$1/4$	$1/16$	$9/16$
3	27	$1/8$	$1/64$	$27/64$
4	81	$1/16$	$1/256$	$81/256$

The number of triangles triples, but each new triangle has a quarter of the area of its parent, because both its base and its height are halved. So the total area is multiplied by $3/4$ at each stage. After 16 stages a little more than one percent of the area is left, since $(3/4)^{16} \approx 0.01002$ , and after 17 stages less than one percent, since $(3/4)^{17} \approx 0.0075$ . Even so, the three sides of the original triangle and the three edges of the first removed triangle lie in every stage, so they all belong to $S$ .

Sierpinski gasket

Definition Sierpinski gasket is another name for the Sierpinski triangle $S$ , and like that name it means the limit set, not a finite stage. The gasket can also be described without removing anything, as three copies of itself. Name the corners $A$ , $B$ and $C$ , and for each corner $v$ let $f_v$ be the map that moves every point of the plane halfway toward $v$ , so that $f_v(p)$ is the midpoint of the segment from $p$ to $v$ . It shrinks every figure to half its size toward that corner. In coordinates, for the corner $v = (a, b)$ ,

f_v(x, y) = \left(\frac{x + a}{2}, \frac{y + b}{2}\right).

Each stage is the union of three half-size copies of the stage before, meaning the set of points that lie in at least one copy, and the gasket is the union of three half-size copies of itself,

\begin{aligned} S_{n+1} &= f_A(S_n) \cup f_B(S_n) \\[0.4em] &\quad \cup f_C(S_n), \\[0.4em] S &= f_A(S) \cup f_B(S) \\[0.4em] &\quad \cup f_C(S). \end{aligned}

The copy $f_v(S)$ is exactly the part of $S$ in the corner triangle at $v$ , and any two of the copies meet in a single point, the midpoint of a side of the original triangle. Call a set closed if it contains every point that its points come arbitrarily close to, and bounded if it fits inside some circle. Among the closed and bounded sets that are not empty, the gasket is the only one equal to the union of its three half-size copies, as the page on iterated function systems explains.

In plain words Instead of cutting holes, you can describe the gasket by shrinking the whole of it to half size three times and placing the copies in the three corners. Each copy carries every hole of the whole, at half the size, which is why each corner has its own middle hole, and each of its corners has one too, at every smaller scale.

Example Use the corners $A = (0, 0)$ , $B = (1, 0)$ and $C = (\tfrac12, \tfrac{\sqrt3}{2})$ , so the side is $1$ . The map toward $A$ is $f_A(x, y) = (x/2, y/2)$ . It takes the three triangles of stage 1 to three triangles of side $\tfrac14$ in the corner at $A$ , and the maps toward $B$ and $C$ do the same in their corners, which gives the $9$ triangles of stage 2. Holes go to holes as well. The center of the original triangle,

G = \left(\tfrac12, \tfrac{\sqrt3}{6}\right),

is also the center of the first removed triangle, so it is not in $S$ . Its image $f_A(G) = (\tfrac14, \tfrac{\sqrt3}{12})$ is the center of the triangle removed from the corner at $A$ at stage 2, so it is not in $S$ either. In general, a point $p$ of the filled triangle $S_0$ belongs to $S$ exactly when $f_A(p)$ does. The point $f_A(p)$ lies in the corner triangle at $A$ , the part of $S$ there is exactly the copy $f_A(S)$ , and $f_A$ never sends two different points to the same place.

Recursive removal

Definition A construction by recursive removal starts with a shape and a rule that removes part of it and leaves smaller pieces of the same kind. The same rule is then applied to every piece that is left, and again to every piece those leave, without end. Each stage contains the next one, since a stage only loses points, and the limit set is made of the points that are never removed, the points common to all the stages. For the Sierpinski triangle the rule removes the open middle triangle, which is its inside without its edges, and leaves three closed corner triangles, which include their edges. Only kept pieces are cut again. A removed middle triangle has no children, so stage $n$ has $3^n$ triangles, not $4^n$ , and the holes cut so far are

1 + 3 + \cdots + 3^{n-1} = \frac{3^n - 1}{2}

removed triangles of different sizes.

In plain words Recursive removal means that one cutting rule is used on every kept piece at every scale. A point survives exactly when it never lies inside a piece that is cut out.

Example By stage 2 the construction has removed one triangle of area $A_0/4$ at stage 1 and three triangles of area $A_0/16$ at stage 2, so the removed area is

\frac{A_0}{4} + \frac{3A_0}{16} = \frac{7A_0}{16},

which matches the $9A_0/16$ that remains. Counting all four triangles of each split as survivors would give $16$ triangles at stage 2 with total area $A_0$ , as if nothing had been removed. In general the area removed in the first $n$ stages is

\begin{aligned} &\frac{A_0}{4}\left(1 + \tfrac34 + \cdots + \left(\tfrac34\right)^{n-1}\right) \\[0.4em] &\quad = A_0 \left(1 - \left(\tfrac34\right)^n\right), \end{aligned}

by the formula for a finite geometric sum, and it approaches the whole area $A_0$ as $n$ grows.

Zero area

Definition A set in the plane has zero area when, for every positive number $\varepsilon$ , it can be covered by a finite or infinite list of triangles whose areas add up to less than $\varepsilon$ . The Sierpinski triangle has zero area. It lies inside every stage $S_n$ , which is covered by $3^n$ triangles of total area $A_0 (3/4)^n$ , and this amount becomes smaller than any positive number as $n$ grows, because every stage multiplies it by $3/4$ . Two cautions go with this. Every finite stage has positive area, so zero area belongs only to the limit set $S$ , and a drawing of a stage with filled triangles shows a set with positive area. And zero area says nothing about whether a set is empty, short or in one piece. A segment has zero area, and the Sierpinski triangle contains infinitely many segments.

In plain words Area measures how much of the plane a set fills. A set of zero area can be trapped inside pieces of total area as small as you like, but it can still contain many points, long paths and a complete outline.

Example Take the original side to be $1$ . The edges of the $3^n$ triangles of stage $n$ all lie in $S$ . Each triangle has perimeter $3 \cdot 2^{-n}$ , and two triangles meet at most at a corner, so these edges have total length

3^n \cdot 3 \cdot 2^{-n} = 3 \left(\tfrac32\right)^n,

which is $3$ , $4.5$ , $6.75$ and $10.125$ for $n = 0, 1, 2, 3$ and grows without bound. So $S$ has zero area but contains paths as long as you like. It also holds together. Every point of $S$ lies in a chain of nested triangles, one from each stage, each a corner triangle of the one before. Start at a corner of the original triangle and walk along an edge to the corner that the original triangle shares with the stage 1 triangle of the chain, then along an edge of that triangle to the corner it shares with the stage 2 triangle, and so on. The steps have lengths at most $1$ , $\tfrac12$ , $\tfrac14$ and so on, so the walk reaches the point in the limit along a path of length at most $2$ that never leaves $S$ . Compare the Cantor set, which is also built by recursive removal and also has zero total length on its line. It contains no interval at all, so no two of its points can be joined by a path inside it. A similar construction does not guarantee a similar result.

Check yourself

A filled triangle has area 256 square centimeters. After four stages, how many filled triangles are left, what is the area of each, and what is the total area? After how many stages is the total area first less than 1 square centimeter?

Show the answer

Each stage triples the number of triangles and divides the area of each one by $4$ . After four stages there are $3^4 = 81$ triangles, each of area $256/4^4 = 1$ square centimeter, so the total is $81$ square centimeters. The formula gives the same total,

256 \cdot \left(\tfrac34\right)^4 = 81.

The total after $n$ stages is $256 \cdot (3/4)^n$ , which decreases at every stage. It is about $1.08$ square centimeters after $19$ stages and about $0.81$ after $20$ , so the total first drops below 1 square centimeter after $20$ stages. By then there are $3^{20} = 3{,}486{,}784{,}401$ triangles, and the set that remains in the limit still has zero area.

Change one detail of the rule. At each stage remove the closed middle triangle, edges included. What happens to the area at each stage, and does what remains still hold together in one piece?

Show the answer

The area does not change. Stage $n$ still has area $A_0 (3/4)^n$ , because the edges of a triangle are segments, which have zero area, and the limit still has zero area.

The connections are lost. At stage 1 the three corner triangles touch only at the midpoints of the sides, and those midpoints are corners of the middle triangle, so the new rule removes them, along with the inner edge of each corner triangle. The three pieces that are left share no point. A path from one corner piece to another would have to pass through a point shared by two corner triangles, and the only such points were removed. The same split happens inside each piece at every later stage, so what remains falls apart into separate parts at every scale. This is why the rule removes only the inside of the middle triangle.

Change the cutting rule. Cut each side of a kept triangle into three equal parts and draw the lines through the cut points parallel to the sides. This divides the triangle into nine small triangles, six pointing the same way as the big one and three pointing the other way. Remove the insides of the three that point the other way. How many triangles are there at stage $n$ , what is their total area, and does it still shrink to zero?

Show the answer

Each small triangle has one third of the base and one third of the height of the big one, so its area is $A_0/9$ . Six of the nine are kept, so each stage keeps $6/9 = 2/3$ of the area. At stage $n$ there are $6^n$ triangles, each of area $A_0/9^n$ , with total area

6^n \cdot \frac{A_0}{9^n} = A_0 \left(\tfrac23\right)^n.

For example, stage 3 has $216$ triangles with total area $8A_0/27 \approx 0.30 A_0$ , compared with $27A_0/64 \approx 0.42 A_0$ for the Sierpinski triangle. Since $2/3$ is less than $1$ , the area still shrinks to zero, and faster than before. It falls below one percent of $A_0$ at stage 12, since $(2/3)^{11} \approx 0.0116$ and $(2/3)^{12} \approx 0.0077$ , compared with stage 17 for the Sierpinski triangle. The six kept triangles cover the three sides of the big triangle and touch one another at corners, so the outline survives and the limit set again holds together in one piece, with zero area.

Sierpinski's Triangle, Holes Inside Holes

Key terms

Sierpinski triangle

Sierpinski gasket

Recursive removal

Zero area

Check yourself

How this lesson connects

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