Prove that $\inf(\{A, B\}) = A \cap B$ Say we have the partial order $\subseteq$ over the power set $P(X)$ from the set X. Prove that $\inf(\{A, B\}) = A \cap B$.
So I've started with proving that $\inf(\{A, B\}) \subseteq A \cap B$ as follows:
Take $M = \inf(\{A, B\})$, then by definition $M \subseteq A$ and $M \subseteq B$, so $M \subseteq (A \cap B)$, therefore $\inf(\{A, B\}) \subseteq (A \cap B)$. Is this part of the proof correct? And I could use some help on finding a proof for $A \cap B \subseteq \inf(\{A, B\})$. Thanks.

You can do a little bit better then that: You don't need to assume additionally, that $\inf(A,B)$ already exists. Simply show that $A\cap B$ has the property of being an infimum:
Certainly $A\cap B \subseteq A$ and $A\cap B \subseteq B$, and if $C$ is another set with $C \subseteq A$ and $C\subseteq B$ then for all $x\in C$ we have $x\in A$ and $x\in B$, hence $x\in A\cap B$. Therefore $C\subseteq A\cap B$. This proves that $A\cap B$ is an infimum of $A$ and $B$.
