# Thread:

1. ## simple proof

How would you prove this, I'm stuck. Thanks

Let A and B be sets. Then A union B = A if and only if B is subset of A.

2. First, you can see from the definition of a subset and from the definition of the union of two sets that for any sets A and B, $A\subseteq{A\cup{B}}$ and $B\subseteq{A\cup{B}}$

Now, to prove our "if and only if" statement, we break it up into the two direction if-then statements:

1. If $A\cup{B}=A$ then $B\subseteq{A}$.
As noted above, $B\subseteq{A\cup{B}}$ for any sets A and B. Thus, if $A\cup{B}=A$, then $B\subseteq{A}$.

2. If $B\subseteq{A}$ then $A\cup{B}=A$.
Assume opposite: we have $B\subseteq{A}$, $A\cup{B}\ne{A}$. As $A\subseteq{A\cup{B}}$, if $A\cup{B}\ne{A}$, then there must be at least one element x in $A\cup{B}$ not in A. By definition of the union of sets, every element of $A\cup{B}$ must be an element of A or of B (or both). Thus if $x\in{A\cup{B}}$ and $x\notin{A}$, then $x\in{B}$. Thus there must be at least one element in B not in A. But this contradicts $B\subseteq{A}$. Thus, we show that if $B\subseteq{A}$ then $A\cup{B}=A$.

With (1) and (2), we have proved that $A\cup{B}=A$ if and only if $B\subseteq{A}$.

-Kevin C.
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