set theory questions [closed]

Question

I need to prove the following:

Given cardinal numbers $ k_1 , k_2, \lambda _1, \lambda_2$

such that $ k_1 \leq k_2 $ and $ \lambda_1 \leq \lambda_2$

Prove

$k_1 + \lambda_1 \leq k_2 + \lambda_2$

I took the definition from my book that said $k + \lambda = |A \cup B| $ where $A$ and $B$ are sets such that $|A|=k$, $|B|=\lambda$ and $A\cap B=\emptyset$

and figured I would have to prove :

Given sets $A$ , $B$, $C$ , $D$

Prove:

a) $A \subseteq A\cup B$

b) if $A \subseteq C$ and $B \subseteq D$ then $A \cup B \subseteq C \cup D$

Answer 1

"A is a subset of B": $A\subseteq B$, formatted $A \subseteq B$.

$(a)$ $A \subseteq A\cup B$.

This means that if $x \in A$, then $x \in A$ OR $x\in B$. This is necessarily true. Every element that's in $A$ is also in the union of $A$ and $B$, since the union of two sets $A, B$ contains all elements that are in $A$, together with all elements that are in $B$.

$(b)$ If $A\subseteq C$ and $B\subseteq D$, then $A\cup B \subseteq C\cup D$.

If $x \in A$, then by the definition of $A\subseteq C$, $x \in C$. Likewise, if $x \in B$, then $x \in D$, for the same reason, since $B \subseteq D$. So if $A \subseteq C$ and $B\subseteq D$, then $x \in A \cup B$ means $x \in A$ or $x \in B$. But we've seen that this means $x \in C$ or $x \in D$, that is, we have that $x \in C\cup D$.