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Look at your and my second answer to the original question. Take your map from $\mathcal{P}(\omega)$ into $\mathcal{L}$ (or rather the set of functions before identifying almost equal elements). That …

I think you are asking too much. Assume we have such a function and let $a$ be nonzero such that both $a_0$ and $a_1$ are nonzero. Then $b\le a_0$ implies $b_1=0$ and $b\le a_1$ implies $b_0=0$. If …

I think the problem is slightly more basic. Fremlin has the fully correct $$ D \Vdash \check D\in\dot{\mathcal{G}} $$ Also, Fremlin did not fix one generic $G$ at the outset; he works with names and t …

The positive solution uses an equivalent of the Axiom of Choice: for every infinite set $A$ there is a bijection $f:A\to A\times A$. In the basic Fraenkel Model (section 4.3 in Jech's Axiom of Choice …

Conditions 4 and 5 show that $(\mathfrak{D}(P),{\subseteq})$ satisfies condition 2 in the list: because $V\in\mathfrak{D}(P)$ we have the second part of 2; and 5 says that $(\mathfrak{D}(P),{\subseteq …

No, $\mathrm{RO}(X)$ is complete; $\mathcal{P}(\omega)/\mathit{fin}$ is not (no strictly increasing sequence has a supremum).

The answer is no: if $\lambda$ is larger than $\omega^2$ and if $X$ contains $\lambda+\omega$ then it also contains $\lambda+\omega+\omega$. To see this observe that $\lambda+1$ is homeomorphic with $ …

Let $M$ be the ordered Mostowski model (T. Jech, The Axiom of Choice, Section 4.5). Its set of atoms, $A$, has a linear order $\prec$ that makes it isomorphic to the rationals. Let $S\in M$ be a subse …

The axiom of choice implies that for every partial order $P$ the hypergraph $H_P$ has property $B$. Let $(P,\le)$ be a partial order. We first claim the following: for every $p\in P$ there is a $q\le …

Here is an attempt at a 'definitive summary'. To begin with positive results: $\mathsf{CH}$ implies a “yes” answer to this question. The fastest way to see this is to first embed a given partial order …