A partial order $\le$ is reflexive, transitive and antisymmetric. It is directed, if for all $x, y$ in the set there is $z$ such that $x \le z$ and $y \le z$.
Let $X, Y$ be partially ordered, directed sets.
A subset $T$ of $X$ is called cofinal (in $X$), if
$\forall x \in X \space \exists t \in T \space (x \le t)$.
$\text{cof}(X) := \text{min} \{|T|: T \text{ is cofinal in } X\}$.
$X, Y$ are called cofinally similar, if there is a partially ordered set $Z$ in which both X and Y can be embedded as cofinal subsets.
In this case, $\text{cof}(X) = \text{cof}(Y)$.
Of course, if $T$ is a cofinal subset of $X$, then $X, T$ are cofinally similar.
This is a well-treated concept, see for instance here and the references given in this paper.
It is relevant for many other branches, for instance topology or the theory of ultrafilters.
This paper mentions an equivalent statement to cofinal similarity in terms of certain maps. This immediately implies that cofinal similarity is transitive.
Now let's define:
$X, Y$ are strongly cofinally similar, if there is a partially ordered set $T$, such that $T$ can be embedded into both, $X$ and $Y$, as a cofinal subset.
Obviously, if $X, Y$ are strongly cofinally similar, they are cofinally similar.
This gives rise to the following questions:
- Does the converse hold, i.e., if $X,Y$ are cofinally similar, are they strongly cofinally similar?
- (formerly question 3) Does the converse hold in certain classes, at least?
Notes
- Of course, 2. is only relevant, if 1. is false.
- A partial answer to 2. is: yes, if $X$ is cofinally similar to a linearly ordered set.
(Since in this case, the regular cardinal $\text{cof}(X)$ embedds into $X$ and $Y$ as a cofinal subset.) - It should be noted that the answer to 1 is affirmative, if and only if strong cofinal similarity is transitive. Thanks to M W for pointing this out. See his comment below for the simple argument.
- Consistently, each directed, non-empty set of size $\le \aleph_1$ is cofinally similar to one of these five sets, each equipped with its standard (partial) order: $1, \omega, \omega_1, \omega_1 \times \omega, [\omega_1]^{< \omega}$. See the above-mentioned paper for the (very complicated) proof (where it is easy to see that these five ones are pairwise not cofinally similar).
Thus, a reasonable source for a counter-example might be a directed set cofinally similar to
$\omega_1 \times \omega$ or to $[\omega_1]^{< \omega}$.
Update Why am I asking this question?
I'm mainly interested in topology. In this area, it's reasonable to compare neighborhood bases just by their cofinality type (= equivalence classes of cofinal similarity).
Here, of course, we consider the ordering defined by reverse set inclusion.
For instance, any open neighborhood base is strongly cofinally similar to the neighborhood base. Or, more interestingly, if $x$ is a point in a dense subset $D$ of a regular topological space $X$, the open neighborhood bases of $x$ in $D$ and in $X$ are strongly cofinally similar.
P-points or points of first countability are completely characterized by the cofinality type of their neighborhood base.
These statements may already motivate question 1: Is it really necessary to distinguish between cofinal similarity and the strong one?
And: can we use the benefits of transitivity to conclude that the neighborhood bases are strongly cofinally similar in the dense subset mentioned above?
Update 2023-10-04
Finally, I found a counter-example, see below.
Best Answer
Eventually, I found a counter-example:
We need a few (well-known) definitions, which I recall here:
$X$ is always a partially ordered set. $X$ is called:
We need the following order-theoretic theorem, which, I think, is well-known. Therefore, I don't include a proof here:
If $X$ is artinian, each antichain in $X$ is finite and every chain in $X$ is countable, then $X$ is countable.
$\le$ is the usual ordering on $\omega_1$. $\mathbb R$ is always endowed with the usual ordering. $(0,1)$ is the open unit interval in $\mathbb R$. $\omega_1 \times \omega$ and $\omega_1 \times \mathbb R$ are endowed with the product order with respect to the usual orderings on their factors.
$L := \{0\} \cup \{\lambda < \omega_1: \lambda \text{ is limit ordinal}\}$
Now let $\varphi: L \rightarrow (0, 1)$ be an injective map.
We extend $\varphi: \omega_1 \rightarrow \mathbb R$ by $\varphi(\lambda + n) = \varphi(\lambda) + n$ for $\lambda \in L, n \in \omega$.
It is easy to see that $\varphi$ is well-defined and injective.
We define a new partial order $\preceq$ on $\omega_1$ by: $\alpha \preceq \beta \Leftrightarrow \alpha \le \beta \text{ and } \varphi(\alpha) \le \varphi(\beta)$
Then:
PROOF. 1. is obvious. 2. holds, since $\preceq$ is coarser than the usual well-ordering. For 3., let $T$ be a chain in $(\omega_1, \preceq)$. Then by 2., $T$ is well-ordered, hence by 1., $\varphi(T)$ is a well-ordered subset of $\mathbb R$, hence countable. Thus, $T$ is countable.
4. Obviously, $\phi$ is an order-embedding. For cofinality, let $(\alpha, r) \in \omega_1 \times \mathbb R$. Then $(\alpha, r) \le \phi(\lambda + n)$ for $\alpha \le \lambda \in L$ and $r \le n \in \omega$.
5. immediately follows by 4., which implies 6.
And finally:
$(\omega_1, \preceq)$ and $\omega_1 \times \omega$ are cofinally similar, but not strongly cofinally similar.
PROOF. Assume $T$ is a cofinal subset of $(\omega_1, \preceq)$ and can cofinally be embedded into $\omega_1 \times \omega $. Then, since this is the case for $\omega_1 \times \omega $, every antichain of $T$ is finite. Hence, by 2., 3. and the above-mentioned theorem, $T$ is countable, which contradicts 6.
Remarks
This kind of ordering on $\omega_1$ and 3. above are well-known, see for instance the book of R. Fraisse, "Theory of Relations", p. 95.
It is not difficult to see that also each antichain in $(\omega_1, \preceq)$ is countable. Hence, in the above order-theoretic theorem, to assume "each antichain is countable" does not suffice.