Assuming A, B, and C are sets prove or disprove the following: 
*

*((A ⊂ C) ∧ (B ⊂ C)) ⇒ A ∪ B ⊂ C

*((A ⊂ B) ∧ (A ⊂ C)) ⇒ ((B ⊆ C) ∨ (C ⊆ B))


Any help would be appreciated.
 A: (1) element-chasing to prove the relation holds:


*

*Let $A \subset C$ and $B \subset C$. Now, if $x \in A \cup B$, then $x \in A \lor x\in B$. 


*

*If $x \in A, x \in C$ because $A \subset C$. 

*If $x \in B$ then $x \in C$ because $B \subset C$.


*In either case, if $A \subset C \land B \subset C$, we have that $x \in A\cup B \implies x\in C$.
(2) is false and all you need to show this is to create a counterexample. There are many to choose from! Any example which makes the left-side of an implication true, but makes the right side of the implication false, will disprove the implication. For example:


*

*Let $A = \{1\}. B = \{1, 2\}, C = \{1, 3\}$.

*Then $A \subset B \land A \subset B$, but $C\not \subseteq B$ and $B \not \subseteq C$.

*Therefore, it is not necessarily the case that $A\subset B \land A \subset C \implies (B \subseteq C) \lor (C \subseteq B)$.

A: HINTS: 
The first statement is true and is very easily proved by element-chasing. Assume that $A\subseteq C$ and $B\subseteq C$, and let $x$ be an arbitrary element of $A\cup B$; why must $x$ belong to $C$?
The second statement is false, and counterexamples are very easy to construct. You could, for instance, let $A=\varnothing$, so that it’s automatically true that $A\subseteq B$ and $A\subseteq C$ no matter what $B$ and $C$ are, and choose $B$ and $C$ to make the statement on the righthand side of the implication false.
