Skip to main content
added 46 characters in body
Source Link

Any partially ordered set $(X,\le)$ is a category, and on such a category, the product $\times$ boils down to meets (greatest lower bounds) and monads boil down to 'closure operators'.

For example, the well-known transitive (reflexive) closure is such a closure operator where the underlying category is the category $\mathcal{C}$ of graphs for a fixed vertex-set $V$, i.e. the objects of this category are relations $E \subseteq V\times V$ and $E_1\le E_2$ iff $E_1\subseteq E_2$. Then, it's easy to see that taking the transitive closure $TE=E^*$ of a relation is a closure operator, i.e. a monad on this category $T\colon \mathcal{C}\to\mathcal{C}$.

In order to see that this monad is non-strong in general, consider some $V$ with at least three (distinct) elements, $x,y,z\in V$. Let $E = \{(x,y), (y,z)\}$ and $R=\{(x,z)\}$. Then, $$ R\times E = \emptyset $$$$ R\times E = R\cap E = \emptyset $$ and so $T\emptyset = \{(v,v) \mid v\in V\}$$T(R\times E) = T\emptyset = \{(v,v) \mid v\in V\}$. But on the other hand, $$ R\times TE = \{(x,z)\}. $$ and so $R\times TE \not\le T(R\times E)$, which shows that transitive closure is a non-strong monad.

Any partially ordered set $(X,\le)$ is a category, and on such a category, the product $\times$ boils down to meets (greatest lower bounds) and monads boil down to 'closure operators'.

For example, the well-known transitive closure is such a closure operator where the underlying category is the category $\mathcal{C}$ of graphs for a fixed vertex-set $V$, i.e. the objects of this category are relations $E \subseteq V\times V$ and $E_1\le E_2$ iff $E_1\subseteq E_2$. Then, it's easy to see that taking the transitive closure of a relation is a closure operator, i.e. a monad on this category $T\colon \mathcal{C}\to\mathcal{C}$.

In order to see that this monad is non-strong in general, consider some $V$ with at least three (distinct) elements, $x,y,z\in V$. Let $E = \{(x,y), (y,z)\}$ and $R=\{(x,z)\}$. Then, $$ R\times E = \emptyset $$ and so $T\emptyset = \{(v,v) \mid v\in V\}$. But on the other hand, $$ R\times TE = \{(x,z)\}. $$ and so $R\times TE \not\le T(R\times E)$, which shows that transitive closure is a non-strong monad.

Any partially ordered set $(X,\le)$ is a category, and on such a category, the product $\times$ boils down to meets (greatest lower bounds) and monads boil down to 'closure operators'.

For example, the well-known transitive (reflexive) closure is such a closure operator where the underlying category is the category $\mathcal{C}$ of graphs for a fixed vertex-set $V$, i.e. the objects of this category are relations $E \subseteq V\times V$ and $E_1\le E_2$ iff $E_1\subseteq E_2$. Then, it's easy to see that taking the transitive closure $TE=E^*$ of a relation is a closure operator, i.e. a monad on this category $T\colon \mathcal{C}\to\mathcal{C}$.

In order to see that this monad is non-strong in general, consider some $V$ with at least three (distinct) elements, $x,y,z\in V$. Let $E = \{(x,y), (y,z)\}$ and $R=\{(x,z)\}$. Then, $$ R\times E = R\cap E = \emptyset $$ and so $T(R\times E) = T\emptyset = \{(v,v) \mid v\in V\}$. But on the other hand, $$ R\times TE = \{(x,z)\}. $$ and so $R\times TE \not\le T(R\times E)$, which shows that transitive closure is a non-strong monad.

Source Link

Any partially ordered set $(X,\le)$ is a category, and on such a category, the product $\times$ boils down to meets (greatest lower bounds) and monads boil down to 'closure operators'.

For example, the well-known transitive closure is such a closure operator where the underlying category is the category $\mathcal{C}$ of graphs for a fixed vertex-set $V$, i.e. the objects of this category are relations $E \subseteq V\times V$ and $E_1\le E_2$ iff $E_1\subseteq E_2$. Then, it's easy to see that taking the transitive closure of a relation is a closure operator, i.e. a monad on this category $T\colon \mathcal{C}\to\mathcal{C}$.

In order to see that this monad is non-strong in general, consider some $V$ with at least three (distinct) elements, $x,y,z\in V$. Let $E = \{(x,y), (y,z)\}$ and $R=\{(x,z)\}$. Then, $$ R\times E = \emptyset $$ and so $T\emptyset = \{(v,v) \mid v\in V\}$. But on the other hand, $$ R\times TE = \{(x,z)\}. $$ and so $R\times TE \not\le T(R\times E)$, which shows that transitive closure is a non-strong monad.