# Boolean algebra and existence of meets

I have a set of sets S that is closed under the set-theoretic operations of union, intersection, and complement, so it basically forms a Boolean algebra with the join operator being union, the meet operator being intersection, and the complement operator being set complement.

Then, I have three subsets A, B, and C of set S.

I know that if A $\subseteq$ B and A $\subseteq$ C then A $\subseteq$ B $\cap$ C.

I do not want to prove that property/law, I just want to use it in my manuscript, but does it have a name?

I found in wikipedia that it is called "existence of meets" (wiki), so is it correct if I say that "As A $\subseteq$ B and A $\subseteq$ C, it holds that A $\subseteq$ B $\cap$ C by existence of meets." or is there a better suggestion?

• What you're using is not just that the meet of $B$ and $C$ exists but that it is the intersection $B\cap C$. – Andreas Blass Jan 25 '17 at 18:06
• thank you for your answer, it is already known that the meet operator is the set intersection because my set forms a Boolean algebra, I just want to know how to properly reference the property that if A ⊆ B and A ⊆ C then A ⊆ B ∩ C. – hamster Jan 25 '17 at 18:16

$P$. If A $\subseteq$ B and A $\subseteq$ C, then A $\subseteq$ B $\cap$ C
is not by the existence of meets, but this is rather (part of) what the existence of meets means. So, you can refer to this result as (part of) the existence of meets, but you don't use the existence of meets to justify or prove $P$. To prove $P$, you would presumably just use the basic definitions of $\subseteq$ and $\cap$.