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A poset or partially ordered set is a set endowed with a partial order, meaning a binary relation $\leq$ which is reflexive ($x \leq x$ for all $x$), antisymmetric ($x\leq y$ and $y\leq x$ implies $x=y$), and transitive ($x\leq y$ and $y\leq z$ implies $x \leq z$).

6 votes

Surjective order-preserving map $f:{\cal P}(X)\to \text{Part}(X)$

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 …
KP Hart's user avatar
  • 11.4k
1 vote
Accepted

Partial orders on downward closed sets

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 …
KP Hart's user avatar
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7 votes
Accepted

Embedding ordinals with the order topology into connected $T_2$-spaces

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 $ …
KP Hart's user avatar
  • 11.4k
7 votes
Accepted

Can any poset of cardinality $\leq 2^{\aleph_0}$ be embedded in ${\cal P}(\omega)/(\text{fin...

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 …
KP Hart's user avatar
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