I was playing around with Sierpinski carpets (see pretty GPU-produced picture here), and came up with a variation that I have been unable to find mentioned elsewhere. I'm wondering if anyone can tell me an existing name for it, and help me determine where it fits into existing classifications of fractals.

As you probably know,

The construction of the Sierpinski carpet begins with a square. The square is cut into 9 congruent subsquares in a 3-by-3 grid, and the [open] central subsquare is removed. The same procedure is then applied recursively to the remaining 8 subsquares, ad infinitum.

In this variation, instead of merely removing the central subsquare, we recursively apply the same procedure to it "inversely." (Is there a better term for that?) That is, we treat the central hole as if we want to replace it with the complement of [this variant of] a Sierpinski carpet.

So when we recur on the central hole, we put an "anti-hole" (like a "donut hole" without a donut) in the middle of it. We will then recur on the anti-hole using the normal (positive) procedure, and recur on the 8 subsquares surrounding the anti-hole using the inverse procedure.

It looks something like this, in its 5th iteration:

alt text

If you think I should post the code for this, I'm happy to, but I didn't think it was very relevant to the math question.

I've been calling it "Sierpinski tablecloth" since I couldn't find another name for it, and because it reminds me of a lace or crocheted tablecloth design.

Correction: @Rahul pointed out that "the limiting set of the recursive process does not exist." This is because the central point of the unit square must be part of both the tablecloth and its complement, or neither - which is impossible. (Of course we can still do finite iterations of the constructive process without hitting a contradiction.)

To fix this problem, I tried proposing that we define the tablecloth set T (arbitrarily) to include the central point. The central subsquare is replaced with the union of the central point with the scaled-down complement of T. As before, the subsquares around the center are replaced with scaled-down copies of T (and therefore the central point of each of the subsquares is contained in T).

Another update: Unfortunately the same problem is true of any point with an irrational coordinate, and of many rational ones. If we express the coordinates in base 3, as with the Cantor set, a point is a member of T iff the number of 1's in the base-3 expansion of each of its coordinates is even. But this doesn't work for numbers with repeating 1's.

If we tried to fix this by making all "problem points" belong to T, we'd be left with "almost" a solid unit square, minus a "dust" of rational points with an even number of 1's in their base-3 coordinate expansions. Then T would cease to generate interesting fractal pictures (in the limit). Its area would be 1, like the irrationals$^2$. It's not clear to me that T would be different from the set of the irrationals$^2$ in any interesting way.

Mind you, we can still draw pictures of the nth iteration of the tablecloth that are fun to look at, but it loses a good deal of its bite if it doesn't exist as a fractal.

Can you think of a way to save the tablecloth from extinction as a fractal?

The rest of this question was written under the misapprehension that the tablecloth exists (is a well-defined set) in the limit.

Fractal dimension

The traditional carpet has Hausdorff dimension

$log(8)/log(3) = ~1.8928$

derived from the fact that it takes 8 self-similar copies at reduction factor 1/3 to cover itself (see example for Sierpinski triangle). In this vein I figure that the tablecloth takes 9 self-similar copies at reduction factor 1/3 to cover itself (no smaller copy will cover the middle mini-square), so it would have Hausdorff dimension

$log(9)/log(3) = 2$

This would put it in the same company (in terms of Hausdorff dimension) as space-filling curves like the Sierpinski curve, Hilbert, Peano, etc. Does this sound reasonable so far?

I'm not quite comfortable with that, though, because a 1/3-size copy of the tablecloth will not actually cover the whole middle subsquare. In fact I can't see how any finite number of reduced-size copies of the tablecloth would cover the whole middle subsquare. And if we say we need an infinite number, the Hausdorff dimension would be infinite. :-(

As an alternative, I considered multifractals, since we seem to have different scaling behavior occurring in different parts of the attractor. But that's deeper water than I'm prepared to wade into.

  • Can anyone point me to literature on fractals that are defined using complements of reduced-size self-similar sets (as well as the usual reduced-size self-similar sets)? Is that a class of fractals that has been studied? (Update:) Or are such "fractals" doomed to be ill-defined (self-contradictory) by their nature?

Similarly to Hausdorff dimension, the box-counting dimension (using the criterion "if the space were divided up into a grid of boxes of size ε, how many boxes of that scale would contain part of the attractor?") also seems to end up at 2, for much the same reasons. And I'm more comfortable with it because a "box" (as opposed to a reduced copy of the tablecloth) will indeed cover the whole middle subsquare. It seems clear that in the limit, the box-counting dimension converges to 2, because there are no gaps, so every box, at whatever size ε, will always contain part of the fractal.

In terms of area (standard Lebesgue measure), I'm pretty sure that the tablecloth has zero, like the carpet, because it has no intervals (rectangles) that are entirely in the set. Right? (Update: I stand corrected: the tablecloth seems to have area 1/2.)


There are several questions above, so to tie them all together: I'm wondering if this tablecloth variant is something new and interesting, or something well-known, or an unknown but trivial variant that can be classified under existing categories. E.g. if there is a variant of the Cantor set that is defined similarly to the tablecloth (for one dimension), I would appreciate a link to info about that.

(Update:) One thing that seems interesting to me about it is that it has points in every interval of the unit square, but it also has holes in every interval. I guess it's like rational numbers in [0,1] in that regard, or like space-filling curves/trees, yet is not a curve or tree (not connected). I can't think of other fractals off-hand that are like that, can you? It seems that this property is connected with the fact that we recur on the inverse instead of leaving holes.

P.S. I just read about the term mutually dense (Interleaving Ad Infinitum): 'If a set S with linear ordering O can be partitioned into two subsets A and B such that between every two distinct elements of A is an element of B, and vice versa, then the sets A and B are said to be "mutually dense".' So the tablecloth and its complement are mutually dense, assuming that this term can be generalized to the unit square. At least in any horizontal or vertical linear slice of the unit square, they are mutually dense.

  • $\begingroup$ So, the complement of the Sierpinski carpet, then? :) $\endgroup$ Oct 21, 2010 at 15:02
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    $\begingroup$ The reason I haven't engaged with your question yet is because I'm still thinking about it :) It's a nice question. $\endgroup$
    – user856
    Oct 21, 2010 at 15:10
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    $\begingroup$ @LarsH: Lebesgue measure is more subtle than that. There exist subsets of the plane of positive Lebesgue measure which contain no rectangles (see, for example, en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set). You should think of the definition of measure zero as being about (the total size of) rectangles covering a set rather than rectangles contained in it. $\endgroup$ Oct 21, 2010 at 15:16
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    $\begingroup$ @LarsH Since the original Sierpinski carpet doesn't have a problem of definition, that object, and presumably scaled copies of it, are fine to use. So if in your construction, instead of doing anything explicitly recursive in the "holes", you just remove an appropriately scaled Sierpinski carpet, you should be fine. In other words use the construction algorithm for the Sierpinski carpet, but whenever a "square" is removed, remove a Sierpinski carpet instead. Seems like it should look similar to what you're after. $\endgroup$
    – yasmar
    Oct 22, 2010 at 14:22
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    $\begingroup$ I don't see why the problem applies to all irrational coordinates - an irrational number can have finitely many $1$s in its ternary expansion. For example, $0.22020020002\dots_3$. $\endgroup$ Feb 1, 2018 at 23:04

4 Answers 4


I wrote a bunch of stuff about the properties of this set assuming it existed, then I realized that it wasn't even obvious that the set was well-defined.

Here's the problem: Does the central point belong to the set? If it does, then it does not belong to the middle block, and vice versa, leading to a contradiction. So the limiting set of the recursive process does not exist.

However, looking at the picture, it feels as though the limit "should" exist in some expanded sense. This could possibly be related to the way that some sequences of functions which do not converge to functions do actually converge to distributions. Or I could just be wrong and the limit simply cannot exist in any sense. Edit: Well, it turns out that the limiting measure contained in any rectangular region is just half the area of the rectangle. So I suppose it does have a limit in the sense of a measure on $[0,1]^2$, but it's not a very interesting measure.

The obvious corresponding version of the Cantor set would be to replace every middle third with the complement of the set, rather than leaving it empty. However, the Cantor set has a very nice interpretation in terms of ternary expansions: a real number $x \in [0,1]$ is in the Cantor set if and only if its ternary expansion contains no digit 1 (allowing, for example, $1/3 = 0.1_3$ to be $0.0222\ldots_3$). For this variant, the corresponding condition would have to be that the ternary expansion contains an even number of ones. This is well-defined for fractions of the form $a/3^b$ whose ternary expansion terminates, but like above, it is not clear how to deal with rationals like $1/2$, or all the irrationals.

  • $\begingroup$ @Rahul: hmm, I had not thought about whether the tablecloth set definition (and by analogy, the corresponding variation on the Cantor set) might be self-contradictory. I wonder if there's an elegant way to "rescue" it. If we said by fiat that the center is a member of the set, and this overrides the other rules, then we have to do that for all reduced-size copies... a countable number of them I believe. But I don't know what this does to fractal dimensions. Probably nothing... $\endgroup$
    – LarsH
    Oct 21, 2010 at 18:29
  • $\begingroup$ @Rahul: My only excuse is that I did say "if the area exists". Good point about needing an even number of 1's in the expansion. You have convinced me the set doesn't exist. $\endgroup$ Oct 21, 2010 at 19:28
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    $\begingroup$ @Ross: I thought that was very prudent of you! Had I been in your place, I would probably have foregone the "if the area exists" bit too. @LarsH: That doesn't solve the problem. There's still the issue of points with irrational coordinates: most of these will go in and out of the set infinitely often as you recurse, so it's not clear how you can say whether any such point remains in the set "in the end". $\endgroup$
    – user856
    Oct 21, 2010 at 22:56
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    $\begingroup$ Yes, I believe that is a proof that the set does not exist. As Rahul said, essentially all irrational points will go in and out of the set infinitely many times during the construction. So it is not well defined whether they are in or out. The central one is just the easiest to define that has this problem. $\endgroup$ Oct 21, 2010 at 23:01
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    $\begingroup$ Once each finite approximation is defined, the “limit” used could be the $\liminf$ or $\limsup$ of this sequence of sets (see wikipedia for definitions). These are always well-defined, and if the sets in the sequence are measurable, they again will be (in any $\sigma$-algebra). They correspond respectively to either including or excluding all of the moot points. $\endgroup$ Jan 11, 2011 at 7:07

The area, if it exists, is $\frac12$. Let the area inside the unit square be A. Then it is made up of $8$ copies at $\frac13$ linear size, plus the complement of a $\frac13$ linear size copy (the center square). So $A=\frac89A+\frac{1-A}{9}$. There is no problem having an area without a rectangle. See the Fat Cantor Set. I believe having positive area also supports dimension $2$.

  • $\begingroup$ @Qiaochu: thanks -- I had read about the Fat Cantor Set but hadn't applied its lesson to this set. So it sounds like measure is closely related to the formula for box-counting dimension. @Ross, cool, I hadn't thought of solving for area that way. $\endgroup$
    – LarsH
    Oct 21, 2010 at 16:02
  • $\begingroup$ Years later, I'm still not sure how the $\frac12$ area makes sense intuitively. I understand the math of solving for $A$, but my spatial intuition wants the area to be > $\frac12$. Maybe just because it starts out at $\frac89$ and approaches $\frac12$ asymptotically, at scales that are hard to see? $\endgroup$
    – LarsH
    Feb 2, 2018 at 16:18
  • $\begingroup$ I think that is correct. The area at each stage is greater than $\frac 12$, but the limit is $\frac 12$ $\endgroup$ Feb 2, 2018 at 16:59

Regarding your update: Using Conway's base 13 function or other constructions (I have one I prefer) you can find a function $f: (0,1) \rightarrow (0,1)$ such that for any interval $(a,b) \subset (0,1)$ and $y \in (0,1)$ there is an $x \in (a,b)$ such that $f(x)=y$. Then for given $y$ the points such that $f(x)=y$ are dense in $(0,1)$. So you have continuum many disjoint sets, each dense in $(0,1)$.

  • $\begingroup$ thanks for this insight. But I need for you to spell out for me how it relates to my update paragraph. I think you're saying that the points $x$ such that $f(x)=y$ (where f is Conway's base 13 function) form an example of a set that, like the rationals, has points in every interval, and also has holes in every interval? $\endgroup$
    – LarsH
    Oct 21, 2010 at 16:51
  • $\begingroup$ Exactly. But even more there are lots of interleaved sets like the rationals, all of which have points in every interval. When we talk about the rationals this way, we usually only think about the single set of the irrationals as the alternative. Vitali's set is another approach, but it doesn't make my head spin the same way. $\endgroup$ Oct 21, 2010 at 16:58

Here's how I attempted to find the area of the Sierpinski tablecloth. Please forgive me if it's not mathematically rigorous enough, I'm new to this.

I tried to find the amount of squares over time with an arithmetic formula. We know for every white square in one iteration it will make one opposite and eight similar blocks. In this case the Black spaces will create white spaces in the style of an inverted Sierpinski carpet. This means that the formula for the amount of black squares is virtually identical to the one for white squares, but reversing the roles of black and white. So:

$$\begin{split} W_n &= 8W_{n-1} + B_{n-1}\\ B_n &= 8B_{n-1} + W_{n-1} \end{split}$$

The group starts as one white block with no black space:

$$W_0 = 1, \ \ \ B_0 = 0.$$

Of course each group blocks will get smaller over time because we make it $1/9$-th the size, and

$$T_n \text{(the area) }= 1-\left(\frac{B_n}{9^n}\right).$$

Using the limit as $W_n$ and $B_n$ approach infinity we can see that:

$(W_n/9^n)$ approaches .5 and $(B_n/9^n)$ approaches .5

so as $T_n$ approaches infinity it becomes $1 - .5$ or .5. The area at infinity is .5.

  • $\begingroup$ Thanks for this way of looking at the problem of the area. It's a different way from @Ross's of arriving at the same answer. I would change the last line to say "So as $n$ approaches infinity, $T_n$ approaches $.5$." I don't think we can say what the area at infinity (i.e. infinite $n$) is, because as Rahul pointed out, "the limiting set of the recursive process does not exist." $\endgroup$
    – LarsH
    Feb 2, 2018 at 14:26
  • $\begingroup$ Can you be more explicit about how you found the limit of $(W_n / 9^n)$ as $n$ approaches infinity? $\endgroup$
    – LarsH
    Feb 2, 2018 at 14:29

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