Is there a category $\mathcal{A}$ of topological spaces, saturated with respect to homeomorphism relation, which is maximal with respect to the following property?


For every $X,Y \in \mathcal{A}$, $X$ is homeomorphic to $Y$ if and only if $X$ can be embedded in $Y$ and $Y$ can be embedded in $X$.



A category of topological spaces with certain anti symmetric property
One way of answering is just by applying Zorn's lemma. 


By the Cantor-Schroeder-Bernstein theorem, we really only have to work one cardinality at a time; that is, for each cardinal $\kappa$, find a maximal class/category $C\_\kappa$ of spaces of cardinality $\kappa$ with this property. Once that is done, then take the union $\bigcup\_\kappa C\_\kappa$ over all cardinalities. 


Then, we only have to worry about homeomorphism equivalence classes, and there is only a small set of homeomorphism classes for each cardinality. So we can just choose a small set of representatives, one for each homeomorphism class, and provided we have a subset of those that is maximal wrt this property, we close it up under saturation. (Point of terminology: in category theory, we say that a full subcategory $D \hookrightarrow C$ is *replete* if for any object $x$ of $D$ and isomorphism $x \cong y$ in $C$, we have that $y$ is an object of $C$. So the last step is called taking the repletion.) 


Let $P\_\kappa$ be the set of chosen representatives of cardinality $\kappa$. We have a poset $Q$ whose elements are subsets $S \subseteq P\_\kappa$ that satisfy the property, ordered by inclusion. It is clear that if $C$ is a chain in $Q$, then the union $U$ of all $S \in C$ satisfies the property, for if $X, Y \in U$ and $X, Y$ embed in each other, then we find some $S$ far enough in the chain so that $X, Y \in S$, and then $X \cong Y$ since $S$ satisfies the property. Then, by Zorn's lemma, there is a maximal element $S$ of $Q$. The repletion of the full subcategory of $Top$ whose objects are elements of $S$ gives us our $C\_\kappa$. 


