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Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ of $\omega$ such that

$A_0=A$;

$|A_{\alpha+1}\setminus A_\alpha|<\omega$; and

$|A_\alpha\setminus A_{\alpha+1}|=\omega$

for every $\alpha<\omega_1$?

Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ of subsets of $\omega$ such that

$A_0=A$;

$|A_{\alpha+1}\setminus A_\alpha|<\omega$; and

$|A_\alpha\setminus A_{\alpha+1}|=\omega$

for every $\alpha<\omega_1$?

Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ of distinct subsets of $\omega$ such that  $A_0=A$ and $|A_{\alpha+1}\setminus A_\alpha|<\omega$ for every $\alpha<\omega_1$?

Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ of $\omega$ such that 

$A_0=A$;

$|A_{\alpha+1}\setminus A_\alpha|<\omega$; and

$|A_\alpha\setminus A_{\alpha+1}|=\omega$

for every $\alpha<\omega_1$?

Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ such that $A_0=A$ and $|A_{\alpha+1}\setminus A_\alpha|<\omega$ for every $\alpha<\omega_1$?

Let $A$ be an infinite subset of $\omega$ such that $\omega\setminus A$ is also infinite.

Under the Continuum Hypothesis is there a sequence $(A_\alpha)_{\alpha<\omega_1}$ of distinct subsets of $\omega$ such that $A_0=A$ and $|A_{\alpha+1}\setminus A_\alpha|<\omega$ for every $\alpha<\omega_1$?