@user14111 yep typo, thx!

For the third, consider an injection of $\omega_1$ into $\mathbb R\setminus \mathbb N$. Can you turn this into a decreasing $\omega_1$-chain?

@user14111 it's clear to me why $\alpha \geq \omega_1$ doesn't work, but I cannot find an easy way to prove that for $\alpha < \omega_1$ there's always a strictly decreasing function

@spaceisdarkgreen wouldn't be problematic if I could? (for the same reason - be able to define an injection $\omega_1 \to \mathbb Q$ - I can't have a strictly increasing/decreasing function from $\omega_1$ to $\mathbb R$, I believe the argoument still holds for $\mathbb R \setminus \mathbb N$ (?))

@leluch_l8r4 why should it be strictly increasing/decreasing? This is for problem 3 (which doesn’t even make mention of the ordering of the reals) not problem 1/2

@leluch_l8r4 for the question of showing any countable ordinal will work for 1, it may be easier to show the stronger statement that any countable linear ordering embeds in $\mathbb Q$. This can be done by enumerating the linear order and constructing the embedding recursively, using the denseness of $\mathbb Q$.

@spaceisdarkgreen sorry for the misunderstanding, now the fact that any countable linear ordering embeds in $\mathbb Q$ it was proved by my professor as a corollary of Cantor's isomorphism theorem (basically taking the product of the linear order and $\mathbb Q$ with the lexicographic order), so I think I can assume this fact without proving it again. Having said that, this fact assures that $\forall \alpha < \omega_1$ there exists an order embedding $\alpha \to \mathbb R$, so the problem now is to find a way to convert this strictly increasing map to a strictly decreasing one, correct?

@leluch_l8r4 sure, or just use that the reverse of the ordinal is just as good of a countable order. Or, alternatively, that $\mathbb R$ is isomorphic to its reverse.

@spaceisdarkgreen so, just to know if I understood correctly, in the second case, given $g$ an isomorphism between $(\mathbb R,<)$ $(\mathbb R,<^*)$, e.g. $x \mapsto -x$, where $<^*$ is the reverse of the usual order of reals $<$, and given an immersion of a countable ordinal $f : \alpha \to \mathbb R$, $g \circ f$ is a strictly decreasing function from $\alpha$ to $\mathbb R$, right? 1/2

@spaceisdarkgreen Instead in the first case you're considering, for every ordinal $\alpha<\omega_1$, the linear order $(\alpha,<^*)$, i.e. the same set but with the reverse order, which is a different order type - and so it's not well-orderd -; by the fact above there exists an order embedding $h : (\alpha,<^*) \to \mathbb R$, and the strictly decreasing function we're looking for is $f : (\alpha,<) \to \mathbb R: \gamma \mapsto h(\gamma)$. Everything correct? 2/2

@leluch_l8r4 yes, looks correct

@user14111 yes, that’s a nice proof (I think you need to restrict to positive reals, but any natural choice of $f$ will already do that), and agree it’s easier, at least if you haven’t already been initiated into the back and forth club, which it seems the op thankfully has. (Though the main “strengthening” is relaxing the countable ordinal to an arbitrary countable ordering.)

@user14111 very nice!

@spaceisdarkgreen about your hint for the third question, I am not (yet) seeing a way to do it, may I ask further details? (sorry to bother this much)

@leluch_l8r4 If $g:\omega_1\to \mathbb R\setminus \mathbb N$ is an injection, let $X_\alpha = (\mathbb R\setminus\mathbb N)\setminus\{g(\beta): \beta < \alpha\}.$ Similarly can you show that it can't work for ordinals $> \omega_1$? This answers the analogue for question 1 for the poset in question 3... note that question 2 and the actual question asked in question 3 aren't immediate from characterizing which ordinals can be reverse-embedded.

@leluch_l8r4 (Regarding "not immediate" I don't mean to psych you out here... it's the most important step toward the answer and it's not that hard to get to the with this information. I just wanted to warn you that it's a substantively different question and the answer isn't just "$\aleph_1$".)

@spaceisdarkgreen well you're right, for question 2 now I know that $\aleph_1 \leq |\text{Dec}(\mathbb R)|$ but I should prove the opposite inequality as well

@leluch_l8r4 I think you need to think a little more about it. A better lower bound that has barely anything to do with part 1 would be $2^{\aleph_0},$ since the first element $f(0)$ in the embedding could be any real number. (Remember, we're counting the embeddings here, not the order types of the embeddings). Part 1 is useful for getting an upper bound.

@spaceisdarkgreen Ah yes, it was obvious, if we consider the following map $\mathbb R \to \text{Dec}(\mathbb R) : r \mapsto f_r$, where $f_r : 2 \to \mathbb R : 0 \mapsto r, 1 \mapsto r-1$, it’s an injection, and so $2^{\aleph_0} \leq |\text{Dec}(\mathbb R)|$. On the other hand, I’ve proved above that $\text{Dec}(\mathbb R) \subseteq \bigcup_{\alpha < \omega_1} {\mathbb R}^\alpha$, and $\left| \bigcup_{\alpha < \omega_1} {\mathbb R}^\alpha \right| \leq 2^{\aleph_0}$, so $|\text{Dec}(\mathbb R)| = 2^{\aleph_0}$, right?

Well actually this basically solves the third request once proved that only the ordinals $< \omega_1$ works for having a strictly decreasing function, in fact, if this assumption is true we have $\text{Dec}(S) \subseteq \bigcup_{\alpha < \omega_1} S^{\alpha}$, and, since is easy to prove that $|S| = 2^{\aleph_0}$, we obtain $|\text{Dec}(S)| \leq 2^{\aleph_0}$. For the opposite inequality the function $\mathbb R \setminus \mathbb N \to \text{Dec}(S) : r \mapsto f_r$, 1/2

where $f_r : 2 \to S : 0 \mapsto (\mathbb R \setminus\mathbb N)\setminus \{r\}, 1 \mapsto (\mathbb R \setminus\mathbb N)\setminus \{r,r^2\}$, is injective and so $2^{\aleph_0} \leq |\text{Dec}(S)|$. So it's basically solved, I'm only taking a little more time to understand why the construction of $X_{\alpha}$ in the hint above gives a contradiction. 2/2