We define an equivalence relation on $\mathcal{P}(\omega)$: for $x,y\in\mathcal{P}(\omega)$ we say $$x\simeq_{fin} y \text{ iff there is } n \in \omega \text{ such that } x\setminus \{0,\ldots,n\} = y \setminus \{0,\ldots,n\}.$$

The set $\mathcal{P}(\omega)/\simeq_{fin}$ is usually written as $\mathcal{P}(\omega)/fin$. For $[x], [y]\in \mathcal{P}(\omega)/fin$ we say $[x]\leq[y]$ iff there is $n \in \omega$ such that $(x\setminus \{ 0,\ldots,n\}) \subseteq (y \setminus \{0,\ldots,n\})$. It is a routine verification that this relation is well-defined.

Does $[x], [y]\in \mathcal{P}(\omega)/fin$ contain an anti-chain of cardinality $2^{\aleph_0}$?

We define an equivalence relation on $\mathcal{P}(\omega)$: for $x,y\in\mathcal{P}(\omega)$ we say $$x\simeq_{fin} y \text{ iff there is } n \in \omega \text{ such that } x\setminus \{0,\ldots,n\} = y \setminus \{0,\ldots,n\}.$$

The set $\mathcal{P}(\omega)/\simeq_{fin}$ is usually written as $\mathcal{P}(\omega)/fin$. For $[x], [y]\in \mathcal{P}(\omega)/fin$ we say $[x]\leq[y]$ iff there is $n \in \omega$ such that $(x\setminus \{ 0,\ldots,n\}) \subseteq (y \setminus \{0,\ldots,n\})$. It is a routine verification that this relation is well-defined.

Does $\mathcal{P}(\omega)/fin$ contain an anti-chain of cardinality $2^{\aleph_0}$?