# simple proof

• Feb 17th 2008, 12:28 PM
rodemich
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.
• Feb 17th 2008, 04:15 PM
TwistedOne151
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, $\displaystyle A\subseteq{A\cup{B}}$ and $\displaystyle B\subseteq{A\cup{B}}$

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

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

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

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

-Kevin C.