# Does a set $A \subseteq [0,1]$ exist such that $A$ is homeomorphic to $[0,1] \setminus A$?

Does a set $A \subseteq [0,1]$ exist such that $A$ is homeomorphic to $[0,1] \setminus A$?

I have no idea how to attack this problem. Any help will be appreciated.

• @snarski $(1 / 2, 1]$ isn't homeomorphic to $[0,1/2]$ Oct 10, 2014 at 5:23
• Do we require the use of the subspace topology on $A$ and $[0,1]\setminus A$? ;) Oct 10, 2014 at 5:24
• Partition of $[0,1]$ to $[0,.)\underline {[.,.)} [.,.] \underline {(.,1]}$ where $A=[0,.)\cup [.,.]$. The intervals should be such that would picture the wagons of a train to move continusly.
– Haha
Oct 10, 2014 at 9:56
• If we take A = all binary expansions in [0,1] with first digit 0 then [0, 1]\A will be all binary expansions with leading digit 1. Are those two sets homeomorphic under the map which exchanges that first digit from 0 to 1?
– Paul
Oct 10, 2014 at 10:32
• @Paul: You have a problem with the dyadic rationals; e.g., $\frac12=0.0111\ldots=0.1$. Oct 10, 2014 at 16:54

Such subsets do exist. The interval $[0,1]$ is homeomorphic to the extended real line $X=[-\infty, \infty]$ (with the standard topology). Now, let $A\subset X$ be the union of $\infty$ with the collection of intervals $[2n, 2n+1)$, $n\in {\mathbb Z}$. The set $A$ is homeomorphic to $B$, which is the union of $\infty$ and the collection of intervals $(2n, 2n+1]$: Send $\infty$ to itself and $[2n, 2n+1)\to (2n, 2n+1]$, $\forall n$ via linear maps. Composing this homeomorphism with the map $x\mapsto -x+1$, we get a homeomorphism $A\to X\setminus A$. qed

Edit: Here is a proof that for such an example ($A$ homeomorphic to $A^{\mathrm c}=[0,1]\setminus A$) the set $A$ has to consist of infinitely many components.

Suppose that $A$ is a finite union of intervals (I allow open, half-open, closed and degenerate intervals). For each interval $I$ define its "modified Euler characteristic" $\chi^c(I)$ as the number of vertices (end-points which belong to $I$) minus the number of edges (which is 1 if $I$ is nondegenerate and $0$ if $I$ is a singleton). Thus, for $I=[a,b]$, we get $\chi^c(I)=1$, while for $I=(0,1)$, we get $\chi^c(I)=-1$; we also have $\chi^c((0,1])=0$.

Now, extend $\chi^c$ to finite unions of intervals in the obvious fashion. For compact subsets with finitely many components, $\chi^c=\chi$, the usual Euler characterstic. One can (easily) show that $\chi^c$ is additive: $$\chi^c(\bigsqcup_{i=1}^n I_i)=\sum_{i=1}^n \chi^c(I_i)$$ (this is false for the usual Euler characteristic!) and is invariant under homeomorphisms.

Now, if $A\subset [0,1]$ is a finite union of intervals and $A^{\mathrm c}$ is homeomorphic to $A$, then $$2\chi^c(A)=\chi^c(A)+ \chi^c(A^{\mathrm c})=\chi^c([0,1])=1$$ which is absurd. The same works for the interval $(0,1)$.

The same argument works in higher dimensions, but you have to modify what "finite number of components" means. Instead, assume that $A$ is "semialgebraic", i.e. is given by a finite system of inequalities of the type $p_i(x)>0$, $p_j(x)\ge 0$, where $p$'s are polynomials of several variables. The key is that $\chi^c$ of the closed $n$-dimensional disk is $1$ and that $\chi^c$ is again additive. The modified Euler characteristic can be regarded as the "right" Euler characteristic for semialgebraic sets; it can be defined as the alternating sum of ranks of homology groups for the Delfs' homology theory ("homology with closed support", not to be confused with Borel-Moore!). This interpretation explains why $\chi^c$ is a topological invariant (this is no longer obvious with the 2nd definition below).

A more direct definition of $\chi^c$ is to consider "incomplete simplicial complexes" triangulating semialgebraic sets, i.e. generalized simplicial complexes where where simplices might be missing some faces (like the interval $[0,1)$ is missing the vertex $1$) and then use the standard alternating sum of the face numbers, as I did above in the 1-dimensional case.

• @Neal the picture of the wagons of the train that moves continously as i said
– Haha
Oct 10, 2014 at 23:06
• Even more simply, observe that $A$ and $X\setminus A$ are both clearly homeomorphic to $\left([0,1)\times\Bbb Z\right)\cup\{p\}$, where $\{p\}\cup\bigcup_{k\ge n}\left([0,1)\times\{k\}\right)$ is a basic open nbhd of $p$ for each $n\in\Bbb Z$. Oct 10, 2014 at 23:18