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:
- chain, if all elements of $X$ are comparable.
- antichain, if $x \le y$ implies $x = y$.
- artinian, if $X$ contains no strictly decreasing sequence.
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:
- $\varphi: (\omega_1, \preceq) \rightarrow
\mathbb R$ is monotonically increasing.
- $(\omega_1, \preceq)$ is artinian.
- Each chain in $(\omega_1, \preceq)$ is countable.
- $\phi: (\omega_1, \preceq) \rightarrow
\omega_1 \times \mathbb R,
\phi(\alpha) = (\alpha, \varphi(\alpha))$
is an order-embedding and $\phi(\omega_1)$
is cofinal in $\omega_1 \times \mathbb R$.
- $(\omega_1, \preceq)$ is directed and
$(\omega_1, \preceq)$ is cofinally similar to
$\omega_1 \times \mathbb R$, hence also to
$\omega_1 \times \omega$.
- $ \text{cof} (\omega_1, \preceq) = \omega_1$
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.
Best Answer
No, a directed set cannot be empty.
This situation is not like the other ones you name. The point of this definition is that a directed set has a direction (that's why it's called that); it is "going somewhere." For example $(\mathbb{N}, \le)$ as a directed set is "going towards $\infty$." In topology directed sets are used to define nets, which are functions from a directed set to a topological space $X$, and these eneralize sequences (which are nets of shape $(\mathbb{N}, \le)$). They have a notion of convergence which generalizes convergent sequences, which relies on this idea that a directed set is "going somewhere."
The empty set is not going anywhere! It does not have a direction.