detecting cofinality in ordinals Suppose K is an unbounded well ordered set, and suppose K is minimal in the sense that each proper initial segment of K has strictly smaller cardinality.
Suppose J is an unbounded well ordered set of the same cardinality as K.
Must there exist a subset M of K, and an order preserving embedding f:M--->J so that f(M) is cofinal in J?
( i.e. for each j in J there exists m in M so that j is less than f(m)).
( In earlier versions of this question, if we overreach and demand that M=K, Brian and Asaf observed the answer is no, even if each final segment of J has the same cardinality as K)
 A: No: the cardinal $\omega_1$, which as an ordinal is the least ordinal of its cardinality, cannot be embedded cofinally in $\omega_1+\omega$ by any order-preserving function, strict or not, and $\omega_1+\omega$ has the same cardinality as $\omega_1$.
Added: Suppose that $f:\omega_1\to\omega_1+\omega$ is (not necessarily strictly) order-preserving and cofinal. For each $k\in\omega$ there is an $\alpha_k\in\omega_1$ such that $f(\alpha_k)\ge\omega_1+k$. Let $\alpha=\sup\{\alpha_k:k\in\omega\}$; then $\alpha_k\le\alpha$ for each $k\in\omega$, so $\omega_1+k\le f(\alpha_k)\le f(\alpha)$ for each $k\in\omega$, and therefore $$f(\alpha)\ge\sup\{\omega_1+k:k\in\omega\}=\omega_1+\omega\;,$$ which is impossible, since $\omega_1+\omega\notin\omega_1+\omega$.
Added2: The answer to the revised version of the question is still no. For a counterexample let $K=\omega_1$, and let $J$ be the ordinal product $\omega_1\cdot\omega$. Pictorially $J$ is $\omega$ copies of $\omega_1$ strung end to end:
$$\overset{\omega_1}\longrightarrow\overset{\omega_1}\longrightarrow\overset{\omega_1}\longrightarrow\overset{\omega_1}\longrightarrow\dots$$
Each final segment of $J$ contains countably infinitely many copies of $\omega_1$, but the cofinality of $J$ is only $\omega$.
A: Yes. M is built by transfinite recursion as follows.
Fix a bijection h:K--->J.
Let m_1=1, the minimal element of K.
Suppose M_*={m_1 < m_2 ....} has been constructed and the restriction of h to M_* is order preserving. 
1) Stop if the restriction has cofinal image and let M=M_*.
2) Otherwise obtain m_* > M_* minimal so that h(m_star) > h(M_star).
To check that the above construction is well defined and eventually terminates in the desired set M, suppose f(M_star) is not cofinal, obtain y>f(M_star), let h(x)=y, and note by construction x>M_star. If at some stage M_* is cofinal in K let M=M_*. The former sentence ensures 2) cannot happen.
