Partition of $\mathbb R$ in convex subsets/badly ordered sets

Question

Background: These questions come from two different exercises, but since the first is much shorter and of the same kind of one of the others, I preferred to put everything in only one thread. (We work in $\mathsf{ZFC}$.)

Question 1: Find the cardinality of the set $S$ of all the partitions in convex subsets of $\mathbb R$.

My approach: I proved that in general the cardinality of the partitions of a set $X$ is $2^{|X|}$ (I think it should be correct); on the opposite side I only find an injection from $\mathbb R$ to $S$. I think it's more reasonable that $|S| = 2^{\aleph_0}$ but I cannot prove the upper bound.

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Definition: Given a totally ordered set $(X,<)$ we say that $(X,<)$ is badly-orderd if there exists an infinite strictly decreasing sequence $(x_i)_{i \in \omega}$ of elements of $X$ and $\forall a \in X \ \exists j \in \omega \ x_j < a$.

Question 2a: Provide an example of a bad order and an example of an order which is neither well (in the common sense) nor bad on the set of naturals $\mathbb N$.

Question 2b: Find the cardinality of the set $R$ of all the classes of isomorphism of bad orders on $\mathbb N$.

My approach: For the first part in question 2a I took $(\mathbb N,<^*)$, where $<^*$ is the converse of the usual order of natural numbers, and it should work fine. For the second part I cannot come up with a good example, any suggestion?

About question 2b I found that the cardinality of the set of bad orders $R'$ on $\mathbb N$ is $2^{\aleph_0}$, and using the projection to the quotient set of equivalence classes $\pi : R' \to R$, I get $|R| \leq 2^{\aleph_0}$. On the other hand every ordinal in $\omega_1 \setminus \omega$ took with the converse order induces a distinct bad order on $\mathbb N$, so $\aleph_1 \leq |R|$. But I cannot decide which of the two claims is correct.

Edit 1: the cardinality of the set of partitions of an infinite set $X$ is actually $2^{|X|}$ as pointed out in the comments, it was actually a typo I made on paper and copied above

Answer 1

As suggested by @IzaakvanDongen this answer is a brief summary of what came out in the comments, therefore one can find further details on the results I am going to summarize in the comments above.

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Question 1. A simple lower bound of $|S|$ can be made with the following injective map: $$\mathbb R \to S : r \mapsto \{]-\infty,r],]r,+\infty[\}.$$ For the upper bound, we notice that a partition of $\mathbb R$ in convex subsets can contain at most $\aleph_0$ (disjoint) convex subsets with cardinality $>1$, otherwise there'd be an injection from a set with cardinality $> \aleph_0$ to $\mathbb Q$, that is absurd. Now this means that every partition $P \in S$ can uniquely be mapped in an element of $\mathcal P_{\leq \aleph_0}(\{\text{convex subsets of $\mathbb R$}\})$ (and the latter has cardinality $2^{\aleph_0}$). Otherwise we can also have the upper bound with the injective map: $$S \to (((\mathbb R \cup \{\pm \infty\})^2 \times \{0,1\}^2) \cup \{\spadesuit\})^\omega : P \mapsto (a_i)_{i \in \omega} $$

where $(a_i)_{i \in \omega}$ encodes the convex subsets with cardinality $>1$ of $P$, and, when these subsets are in a finite number, the map gives eventually $\spadesuit$ (this needs further details to be written properly, but I think the idea is pretty clear).

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Question 2a. A first remark that's needs to be done is that a set is badly-ordered if and only if doesn't have a minimum element, so we're looking for a countable total order that has a minimum element but at the same time has a non-empty subset without a minimum element.
Two simple examples can be done, the first taking the following subset of rationals: $$\{-1\} \cup \left\{\frac 1n : n \in \omega \setminus \{0\}\right\}$$ the second one taking the sum of total orders: $$(\{\pi\},<) + (\mathbb Q,<)$$ Both examples are countable total orders with a minimum element and a subset without minimum, so taking a bijection with $\mathbb N$, we can induce one of the order above on $\mathbb N$.

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Question 2b. As un upper bound, the one given at the end of the question above is correct, now we provide a way to encode an countable binary sequence in a bad ordered set. We define: $$\{0,1\}^\omega \to R : (a_i)_{i \in \omega} \mapsto A:= \mathbb Z \sqcup \bigsqcup_{i \in \omega} F(a_i)$$ where $F(a_i) = \mathbb Z$ if $a_i = 1$, otherwise $F(a_i) = \{i\}$. This is a countable sum of total orders, so it's a countable total order, and in particular, thanks to the first copy of $\mathbb Z$, it's a bad order. To prove that different sequences give us non isomorphic orders we can prove by induction that if we have $(a_i)_{i \in \omega} \mapsto A \cong B \leftarrow\!\shortmid (b_i)_{i \in \omega}$, then $\forall n \in \omega \ (a_i)|_n = (b_i)|_n$, which implies $(a_i)_{i \in \omega} = (b_i)_{i \in \omega}$. So we conclude that the map is well-defined and injective.