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Let x $\in A \cup B$
Case 1: $x \in A$
{$A: x \le M$}

Case2: $x \in B$
{$B: x \le N$}

Since both A and B are bounded above, there exist a real number K = M + N where
{$A \cup B: x \le K$}, therefore $A \cup B$ must be bounded above by K.

Question
I don't know how to fix the case when either the M or N bound is negative. I am new to set and proofing so any help would be appreciated

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    The notation $\text{max}(a,b):=\begin{cases}a&\text{if}~a\geq b\\ b&\text{if }~b>a\end{cases}$ is useful here. – 2017-02-01
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    So I can say the there exist a $K = max(M.N)$. For case1: $x \le M$ and $x \lt K$ and for case 2: $x \le N$ and $x \lt K$. Becuase $x$ is always less than $K$, then there exist an upper bound where {$A \cup B: x \le K$}. Therefore $A \cup B$ is bounded. – 2017-02-01

1 Answers 1

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Your problem is that you should not be adding $M$ and $N$ to get the upper bound of $A \cup B$. As you say, that can fail when one of $M,N$ is negative. You are right to be looking for a $K$ that bounds them both. In what cases could it be $M$?