Let $\lambda$ be a singular cardinal of countable cofinality.


Is there necessarily a sequence $\{A\_\alpha\mid\alpha<\lambda^+\}$ of countable subsets of $\lambda$, such that $\alpha<\beta$ if and only if $A\_\alpha\setminus A\_\beta$ is finite?


In other words, we know that for $\omega$, there is an uncountable sequence of subsets which is increasing modulo finite changes. Is the same true for other cardinals of countable cofinality?



"Towers" on singular cardinals with countable cofinality
For $\lambda > 2^{\aleph\_0}$, there is no such sequence.


Suppose $\lambda > 2^{\aleph\_0}$. Because $2^{\aleph\_0}$ cannot have countable cofinality, there is some $\kappa < \lambda$ with $2^{\aleph\_0} < \kappa$. Consider the sequence $\{A\_\alpha \cap A\_\kappa \mid \alpha < \kappa\}$. For each particular $\alpha$, the sets $A\_\alpha$ and $A\_\alpha \cap A\_\kappa$ differ by only finitely many elements. Each set of this form is a countable subset of the countable set $A\_\kappa$, so there are fewer than $\kappa$ possibilities for the sets $A\_\alpha \cap A\_\kappa$. Thus we may find $\alpha < \beta < \kappa$ such that $A\_\alpha \cap A\_\kappa = A\_\beta \cap A\_\kappa$. But then $A\_\alpha$ and $A\_\beta$ differ by only finitely many elements, and in particular we have $A\_\beta \setminus A\_\alpha$ finite while $\alpha < \beta$, contrary to your condition.


