2
$\begingroup$

I am currently working through Sam Nadler's "Continuum Theory" and have become stuck on an exercise (1.16) which says "Prove that the Sierpinski Carpet is universal." By universal they means that the Sierpinski Carpet contains a topological copy of every one dimensional plane Continuum. I am interpreting "contains a topological copy" to mean that every one dimensional plane Continuum embeds into the carpet.

Is there a reference for this result that is in English?

$\endgroup$
  • $\begingroup$ Nöbeling space $N_n$ is also universal for all separabel metric spaces of dimension $n$. That is in Engelking's books (both in "general topology", and more fully in his "theory of dimensions, finite and infinite"), e.g. $N_1$ (all points of $[0,1]^3$ with at most one rational coordinate) is quite similar to the Sierpinski carpet. So that proof might give you ideas. $\endgroup$ – Henno Brandsma Sep 17 '17 at 21:47
  • $\begingroup$ Much appreciated. I'll take a look. $\endgroup$ – Robert Thingum Sep 17 '17 at 21:48
  • $\begingroup$ G. Whyburn Topological characterization of the Sierpinski curve, Fundamenta Math. 45 (1958) 320–324 $\endgroup$ – Moishe Kohan Sep 19 '17 at 3:03
2
$\begingroup$

Well, I was unable to find a (legal) copy of Engelking's book and Whyburn's paper, while suggestive, does not give the proof I was hoping to find. I ended up running Sierpinski's original paper (easily found through Wikipedia) through google translate (which has come a long way it seems) and tried to parse his proof. What follows is my attempt to reproduce at least the idea of what he did originally. Hopefully my TeX formatting works out.

First some preliminary definitions and observations.

Throughout, let $S$ denote the Sierpinski carpet as a subset of $I^{2}=[0,1]\times[0,1]$. Let $R\subseteq\mathbb{R}^{2}$ be a rectangle with vertices $(\alpha_{1},\beta_{1}),(\alpha_{2},\beta_{1}),\alpha_{1},\beta_{2})$ and $(\alpha_{2},\beta_{2})$ with $\alpha_{1}<\alpha_{2}$ and $\beta_{1}<\beta_{2}$. Then, because $K$ is both closed and $1$ dimensional $K$ is not dense in $R$, if it intersects $R$ at all, so there is a rectangle $R^{\prime}\subseteq R$ such that $R^{\prime}\subseteq int(R)$ and $int(R^{\prime})\cap K=\emptyset$. Say a rectangle $R$ has property $P$ if its side are parallel to the coordinate axes and its interior is disjoint from $K$. Then define $P(R)$ to be the set of all rectangles in $int(R)$ that have property $P$. Likewise define $X(R)$ to be the set of all rectangles contained in $R$. Let $f:X(R)\rightarrow P(R)$ be a choice function such that $f(C)\subseteq int(C)$.

Given a rectangle $R$, the rectangle $f(R)$ corresponds to a subdivision of $R\setminus int(f(R))$ into $8$ rectangles $R_{1},\ldots,R_{8}$. Specifically, if $R$ has coordinates determined by numbers $\alpha_{1}<\alpha_{2}$ and $\beta_{1}<\beta_{2}$ as in the previous paragraph and $f(R)$ is determined by $x_{1}<x_{2}$ and $y_{1}<y_{2}$ then the rectangles of the subdivision are given by,

$$\begin{array}{rcl} R_{1} & = & [\alpha_{1},x_{1}]\times[\beta_{1},y_{1}]\\ R_{2} & = & [x_{1},x_{2}]\times [\beta_{1},y_{1}]\\ R_{3} & = & [x_{2},\alpha_{2}]\times[\beta_{1},y_{1}]\\ R_{4} & = & [\alpha_{1},x_{1}]\times[y_{1},y_{2}]\\ R_{5} & = & [x_{2},\alpha_{2}]\times[y_{1},y_{2}]\\ R_{6} & = & [\alpha_{1},x_{1}]\times[y_{2},\beta_{2}]\\ R_{7} & = & [x_{1},x_{2}]\times[y_{2},\beta_{2}]\\ R_{8} & = & [x_{2},\alpha_{2}]\times[y_{2},\beta_{2}] \end{array}$$

Pictorially, illustration of a subdivision:

Illustration of a subdivision

We will refer to this subdivision as the subdivision induced by $f(R)$ and the $R_{i}$ as rectangles created in this subdivision.

Now let $K$ be a $1$ dimensional plane continuum. Without loss of generality $K$ is contained in the unit square $I^{2}$. Divide $I^{2}$ into $9$ squares each of area $1$ as in the construction of $S$. Let $V$ be the center square and let $f:X(I^{2})\rightarrow P(I^{2})$ be as described. Then $f(V)$ is well defined. We then define

$$C_{1}=U\setminus int(f(V))$$

We then consider the subdivision induced by $f(V)$. This gives us rectangles $R_{1},\ldots,R_{8}$. We then define $C_{2}\subseteq C_{1}$ to be

$$C_{2}=C_{1}\setminus\bigcup_{i_{1}=1}^{8}int(f(R_{i_{1}}))$$

Then $f(R_{i_{1}})$ induces a subdivision of $R_{i_{1}}$ into $8$ rectangles $R_{i_{1}1},R_{i_{1}2},\ldots,R_{i_{1}8}$. We then define $C_{3}\subseteq C_{2}$ to be

$$C_{3}=C_{2}\setminus\bigcup_{i_{1}=1}^{8}\bigcup_{i_{2}=1}^{8}int(f(R_{i_{1}i_{2}}))$$

and in general

$$C_{n}=C_{n-1}\setminus\bigcup_{i_{1}=1}^{8}\cdots\bigcup_{i_{n-1}=1}^{8}int(f(R_{i_{1}\ldots i_{n-1}}))$$

Then define $Y=\bigcap_{n=1}^{\infty}C_{n}$. This is clearly a subset of $I^{2}$ that contains an embedding of $K$. We claim that $Y$ is homeomorphic to $S$. To see this let $B_{1}$ denote the first stage in the construction of $S$. That is $S=I^{2}\setminus (\frac{1}{3},\frac{2}{3})^{2}$. Then $B_{1}$ is homeomorphic to $C_{1}$. More generally $B_{n}$ is homeomorphic to $C_{n}$. Let $\{f_{n}\}$ be the sequence of these homeomorphisms. Then each $f_{n}$ also provides an embedding of $S$ into $C_{n}$, and in particular into $I^{2}$. This sequence uniformly converges to a function $f$ which, because the sequence converged uniformly, is continuous. Moreover $f(\partial B_{i})=f_{i}(\partial B_{i})$ for all $i$. This $f$ is an embedding of $S$ into $I^{2}$ and is a homeomorphism between $S$ and $Y$ provided that it is injective as a function from $S$ into $I^{2}$.

However, we needn't discuss $f$ because $Y$ must be homeomorphic to $S$ by the main result of Whyburn's paper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.