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substituted the word 'class' for 'type'
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Todd Trimble
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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 typesequivalence classes, and there is only a small set of homeomorphism typesclasses for each cardinality. So we can just choose a small set of representatives, one for each homeomorphism typeclass, 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 $C_\kappa$.

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 types, and there is only a small set of homeomorphism types for each cardinality. So we can just choose a small set of representatives, one for each homeomorphism type, 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 $C_\kappa$.

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 $C_\kappa$.

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Todd Trimble
  • 53.3k
  • 6
  • 205
  • 322

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 types, and there is only a small set of homeomorphism types for each cardinality. So we can just choose a small set of representatives, one for each homeomorphism type, 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 $C_\kappa$.