It is currently Thu Sep 19, 2019 5:02 am

 All times are UTC [ DST ]

 Page 1 of 1 [ 2 posts ]
 Print view Previous topic | Next topic
Author Message
 Posted: Wed Nov 11, 2015 4:18 pm

Joined: Mon Nov 09, 2015 11:52 am
Posts: 77
Location: Limassol/Pyla Cyprus
This is a question I posted in the old forum but remained unanswered. Let us see if it will have a better fate this time.

Decide whether you can find uncountably many disjoint subsets of $\mathbb{R}$ such that each of them is order-similar to $\mathbb{R}.$ (I.e. for each of them there is a bijection with $\mathbb{R}$ which preserves the ordering.)

Top

 Posted: Tue Sep 26, 2017 10:09 am

Joined: Mon Nov 09, 2015 11:52 am
Posts: 77
Location: Limassol/Pyla Cyprus
It remained unanswered for quite some time. Since I have been asked for it, here is a solution:

Given a sequence $a = (a_1,a_2,a_3,\ldots)$, where $a_i \in \{1,2\}$ for each $i$, we construct the subset $S_a$ of $(0,1)$ as follows:

Every element of $S_a$ is of the form $x_b = \frac{1}{2^{b_1}} + \frac{1}{2^{b_1+b_2}} + \cdots$

where each $b_i$ is a positive integer congruent to $a_i$ modulo $2$.

In fact, for reasons that will become clear later, instead of $S_a$ we will consider $S_a' = S_a \setminus \{x_a\}$, where

$x_a = \frac{1}{2^{a_1}} + \frac{1}{2^{a_1+a_2}} + \cdots$

It is easy to check that we have constructed uncountably many disjoint subsets of $\mathbb{R}$. This relies on the fact that each real number has a unique base 2 expansion provided that we disallow terminating expansions.

So it remains to show that each $S_a'$ is order-similar to $\mathbb{R}$. Consider the map

$\frac{1}{2^{b_1}} + \frac{1}{2^{b_1+b_2}} + \cdots \mapsto \frac{1}{2^{\lceil b_1/2 \rceil}} + \frac{1}{2^{\lceil b_1/2 \rceil +\lceil b_2/2 \rceil}} + \cdots$

It is easy to check that this map reverses order. Furthermore, the image set of this map is $(0,1)$. Rather than giving a formal proof it is easier to exhibit it by an example. If $x = 0.0100110\ldots$ in its binary expansion, then we want to choose $b_1$ such that $\lceil b_1/2\rceil = 2$. So we need $b_1=3$ or $b_1=4$. Only one of the two choices is allowed as we need $b_1 \equiv a_1 \bmod 2$. Then, we need $\lceil b_1/2\rceil + \lceil b_2/2\rceil = 5$. This gives a unique choice for $b_2$, and so on.

Here is the place where we use $S_a'$ instead of $S_a$. The element $x_a$ would map to the element $1$ which we want to exclude in order to have an open interval as an image.

Now it remain to find a bijection between $(0,1)$ and $\mathbb{R}$ which again reverses order. This is easy. For example we can take $\arctan(\pi/2 - \pi x)$.

Top

 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 Page 1 of 1 [ 2 posts ]

 All times are UTC [ DST ]

#### Mathimatikoi Online

Users browsing this forum: No registered users and 1 guest

 You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forumYou cannot post attachments in this forum

Search for:
 Jump to:  Select a forum ------------------ Algebra    Linear Algebra    Algebraic Structures    Homological Algebra Analysis    Real Analysis    Complex Analysis    Calculus    Multivariate Calculus    Functional Analysis    Measure and Integration Theory Geometry    Euclidean Geometry    Analytic Geometry    Projective Geometry, Solid Geometry    Differential Geometry Topology    General Topology    Algebraic Topology Category theory Algebraic Geometry Number theory Differential Equations    ODE    PDE Probability & Statistics Combinatorics General Mathematics Foundation Competitions Archives LaTeX    LaTeX & Mathjax    LaTeX code testings Meta