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 ordersimilar to \(\mathbb{R}.\) (I.e. for each of them there is a bijection with \(\mathbb{R}\) which preserves the ordering.)
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Uncountably many disjoint subsets similar to \(\mathbb{R}\)?

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Re: Uncountably many disjoint subsets similar to \(\mathbb{R}\)?
It remained unanswered for quite some time. Since I have been asked for it, here is a solution:
The answer is affirmative.
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 ordersimilar 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 $\arctan(\pi/2  \pi x)$.
The answer is affirmative.
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 ordersimilar 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 $\arctan(\pi/2  \pi x)$.