A set $A\subseteq \omega$ is said to be thin if $$\lim\sup_{n\to\infty}\frac{|A\cap \{0,\ldots, n\}|}{n+1} = 0.$$

We say for $A, B\subseteq \omega$ that $A\simeq_\text{fin} B$ if the symmetric difference $A \, \triangle \, B$ is finite. It is easy to see that this an equivalence relation, and a routine verification shows that the resulting quotient is an atomless Boolean algebra. It is customarily denoted by ${\cal P}(\omega)/(\text{fin})$.

We can do exactly the same thing for thin sets by saying $A\simeq_{\text{thin}} B$ if $A \, \triangle \, B$ is thin. This gives rise to a Boolean algebra ${\cal P}(\omega)/(\text{thin})$.

Question. Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic? ${}$

A set $A\subseteq \omega$ is said to be thin if $$\lim\sup_{n\to\infty}\frac{|A\cap \{0,\ldots, n\}|}{n+1} = 0.$$

We say for $A, B\subseteq \omega$ that $A\simeq_\text{fin} B$ if the symmetric difference $A \, \triangle \, B$ is finite. It is easy to see that this an equivalence relation, and a routine verification shows that the resulting quotient is an atomless Boolean algebra. It is customarily denoted by ${\cal P}(\omega)/(\text{fin})$.

We can do exactly the same thing for thin sets by saying $A\simeq_{\text{thin}} B$ if $A \, \triangle \, B$ is thin. This gives rise to a Boolean algebra ${\cal P}(\omega)/(\text{thin})$.

Question. Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic?

A set $A\subseteq \omega$ is said to be thin if $$\lim\sup_{n\to\infty}\frac{|A\cap \{0,\ldots n\}|}{n+1} = 0.$$

We say for $A, B\subseteq \omega$ that $A\simeq_{\text{fin}}$ if the symmetric difference $A \, \triangle \, B$ is finite. It is easy to see that this an equivalence relation, and a routine verification shows that the resulting quotient is an atomless Boolean algebra. It is customarily denoted by ${\cal P}(\omega)/(\text{fin})$.

We can do exactly the same thing for thin sets by saying $A\simeq_{\text{thin}} B$ if $A \, \triangle \, B$ is thin. This gives rise to a Boolean algebra ${\cal P}(\omega)/(\text{thin})$.

Question. Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic?

A set $A\subseteq \omega$ is said to be thin if $$\lim\sup_{n\to\infty}\frac{|A\cap \{0,\ldots, n\}|}{n+1} = 0.$$

We say for $A, B\subseteq \omega$ that $A\simeq_\text{fin} B$ if the symmetric difference $A \, \triangle \, B$ is finite. It is easy to see that this an equivalence relation, and a routine verification shows that the resulting quotient is an atomless Boolean algebra. It is customarily denoted by ${\cal P}(\omega)/(\text{fin})$.

We can do exactly the same thing for thin sets by saying $A\simeq_{\text{thin}} B$ if $A \, \triangle \, B$ is thin. This gives rise to a Boolean algebra ${\cal P}(\omega)/(\text{thin})$.

Question. Are the Boolean algebras ${\cal P}(\omega)/(\text{fin})$ and ${\cal P}(\omega)/(\text{thin})$ isomorphic? ${}$