Show that inclusion relation ⊆ is a partial ordering on the power set of a set S. Draw the Hasse diagram for the partial ordering {(A,B) | A ⊆ B} on the power set P(S) where S = {a,b,c}.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

Let us show that inclusion relation $⊆$ is a partial ordering on the power set of a set $S$ .

Since $A\subseteq A$ for any subset $A\subseteq S,$ we conclude that this relation is reflexive.

Taking into account that $A\subseteq B$ and $B\subseteq A$ imply $A=B,$ we conclude that the relation is antisymmetric. Since $A\subseteq B$ and $B\subseteq C$ imply $A\subseteq C$ , it follows that this relation is transitive.

Consequently, the inclusion relation is a partial ordering on the power set of a set $S$ .

Let us draw the Hasse diagram for the partial ordering $\{(A,B) | A ⊆ B\}$ on the power set $P(S)$ where $S = \{a,b,c\}.$

Show that inclusion relation ⊆ is a partial ordering on the power set of a set S. Draw the Hasse diagram for the partial ordering {(A,B) | A ⊆ B} on the power set P(S) where S = {a,b,c}.

Here's the Solution to this Question

Related Answers

Was this answer helpful?

Join our Community to stay in the know

Question ID: mtid-5-stid-8-sqid-953-qpid-808