ThmDex – An index of mathematical definitions, results, and conjectures.
Intersection is largest lower bound
Formulation 0
Let $X$ be a D11: Set.
Let $E_j$ be a D11: Set for each $j \in J$ such that
(i) \begin{equation} J \neq \emptyset \end{equation}
(ii) $\bigcap_{j \in J} E_j$ is the D76: Set intersection of $E = \{ E_j \}_{j \in J}$
Then
(1) \begin{equation} \forall \, i \in J : \bigcap_{j \in J} E_j \subseteq E_i \end{equation}
(2) \begin{equation} \forall \, j \in J : X \subseteq E_j \quad \implies \quad X \subseteq \bigcap_{j \in J} E_j \end{equation}
Subresults
R4148: Intersection is a lower bound to each set in the intersection
Proofs
Proof 0
Let $X$ be a D11: Set.
Let $E_j$ be a D11: Set for each $j \in J$ such that
(i) \begin{equation} J \neq \emptyset \end{equation}
(ii) $\bigcap_{j \in J} E_j$ is the D76: Set intersection of $E = \{ E_j \}_{j \in J}$
Fix $i \in J$ and $x \in \bigcap_{j \in J} E_j$. By definition of set intersection, $x \in E_j$ for every $j \in J$ and thus also for $i$. This proves (1).

Suppose then that $X \subseteq E_j$ for every $j \in J$. If $X$ is empty, then the claim is a consequence of R7: Empty set is subset of every set, so we may assume that $X$ is nonempty and fix $x \in X$. Since $X$ is contained in $E_j$ for each $j \in J$, then $x \in E_j$ for every $j \in J$ and thus, by definition of set intersection, $x \in \bigcap_{j \in J} E_j$. This concludes the proof. $\square$