6
$\begingroup$

Let $A,B$ be non-empty, dense in itself (i.e. every point is a limit point), compact subsets of $\mathbb R$ with empty interior. Then how to show that there exists order preserving isomorphism $f: \mathbb R \to \mathbb R$ such that $f(A)=B$ ?

I can see that both $A,B$ are perfect sets (https://en.wikipedia.org/wiki/Perfect_set) , so has cardinality of continuum. Also, it is enough to take one of $A$ or $B$ to be the Cantor set. I can show that $A, B$ are order homeomorphic with the cantor set, but I don't know if such a homeomorphism can be extended to whole real line.

I am unable to derive anything else.

Please help . Thanks in advance

$\endgroup$
  • $\begingroup$ Once you show the existence of an order-preserving homeo $A\to B$, extend it to the real line linearly on each complementary interval. $\endgroup$ – Moishe Kohan Jan 3 '18 at 16:21
  • $\begingroup$ @MoisheCohen : what do you mean by "extend it to the real line linearly on each complementary interval" ? $\endgroup$ – user495643 Jan 3 '18 at 16:22
  • $\begingroup$ Do you know what a linear function is? $\endgroup$ – Moishe Kohan Jan 3 '18 at 16:23
  • $\begingroup$ @Moishe Cohen : yes, you mean a linear map right ? Of the form $ax+b$ ... but there are some I am seeing ... U guess your plan is to write the complement of $A$ and the cantor set as a unique disjoint union of open intervals, and map the intervals onto each other linearly .... but how do you know that both the complements have same no. of disjoint intervals ? And secondly, patching those linear maps, how is it that the resulting map is order preserving ? Or is your plan something else ... ? $\endgroup$ – user495643 Jan 3 '18 at 16:46
  • $\begingroup$ Correct! ... This "number" is actually countably infinite. Try to prove this yourself, this is a very good exercise. $\endgroup$ – Moishe Kohan Jan 3 '18 at 17:31
2
+25
$\begingroup$

Theorem. (Cantor). Let $X$ be countably infinite and linearly ordered by $<_X$ with no $<_X$- largest nor $<_X$-smallest member, and such that $<_X$ is order-dense . (That is, if $a<_Xa'$ then there exists $a''$ with $a<_Xa''<_X a').$ Then $(X,<_X)$ is order-isomorphic to $(\Bbb Q,<)$ where $<$ is the usual order on $\Bbb Q$.

Corollary. Any two such linear orders are order-isomorphic to each other.

We have $\Bbb R$ \ $A= (-\infty, \min A)\cup (\max A,\infty)\cup (\cup F_A)$ where $F_A$ is a countably infinite family of pair-wise disjoint non-empty bounded open intervals whose closures are pair-wise disjoint.

We have $\Bbb R$ \ $B=(-\infty,\min B)\cup (\max B,\infty)\cup (\cup F_B)$ where $F_B$ is a countably infinite family of pair-wise disjoint non-empty bounded open intervals whose closures are pair-wise disjoint.

For $U,V\in F_A$ let $U<_A V \iff \sup U<\inf V.$

For $U',V'\in F_B$ let $U'<_B V'\iff \sup U'<\inf V'.$

By the corollary above, there is an order-isomorphism $\psi:F_A\to F_B.$ That is, $\psi$ is a bijection and $U<_AV\iff \psi (U)<_B \psi (V).$

For $U\in F_A$ let $f$ map $\overline U$ linearly onto $\overline {\psi(U)}$ with $f (\inf U)=\inf (\psi (U)).$ Let $f$ map $(-\infty,\min A]$ linearly onto $(-\infty,\min B].$ Let $f$ map $[\max A,\infty)$ linearly onto $[\max B,\infty).$

For $x\in A,$ if $x$ is not an end-point of any member of $F_A,$ and if $\min A \ne x \ne \max A$ then $x=\sup \{\sup U: U\in F_A\land \sup U<x\}$. Let $f(x)=\sup \{\sup \psi(U): U\in A\land \sup U<x\}. $

Remark: The corollary can be proven directly by elementary means.

$\endgroup$
  • $\begingroup$ Why is $F_A$ a union of countably "infinitely" many ? I can see ciuntable, but why infinite ? $\endgroup$ – user495643 Jan 7 '18 at 15:44
  • $\begingroup$ $\cup F_A\subset (\min A,\max A).$....If $F_A$ is finite let $x=\min (\{\max A\}\cup \{\inf U: U\in F_A\}).$ If $x=\min A$ then $x$ is an isolated member of $A$ and if $ x>\min A$ then $ (\min A,x)$ is a non-empty open subset of $A.$ $\endgroup$ – DanielWainfleet Jan 7 '18 at 17:31

Your Answer

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