- #1
danago
Gold Member
- 1,123
- 4
Suppose that [tex]A,B \subseteq \Re^+[/tex] are non empty and bounded sets. Define + and . as the following set opetations:
[tex]
\begin{array}{l}
A + B = \{ a + b|a \in A,b \in B\} \\
A.B = \{ ab|a \in A,b \in B\} \\
\end{array}
[/tex]
Prove that [tex]\sup (A + B) \le \sup A + \sup B[/tex]
I started by letting [tex]a \in A,b \in B[/tex]. From the definition of supremum:
[tex]
\begin{array}{l}
a \le \sup A \\
b \le \sup B \\
\end{array}
[/tex]
I then added the two inequalities to give:
[tex]
a + b \le \sup A + \sup B
[/tex]
Since this holds true for any a and b, sup A+sup B is an upper bound on the set A+B. This is where I am a bit stuck. I can't really see why sup(A+B) isn't strictly equal to sup A + sup B. I tried coming up with a few example sets A and B but that didnt really help anything.
Any suggestions?
Thanks,
Dan,
[tex]
\begin{array}{l}
A + B = \{ a + b|a \in A,b \in B\} \\
A.B = \{ ab|a \in A,b \in B\} \\
\end{array}
[/tex]
Prove that [tex]\sup (A + B) \le \sup A + \sup B[/tex]
I started by letting [tex]a \in A,b \in B[/tex]. From the definition of supremum:
[tex]
\begin{array}{l}
a \le \sup A \\
b \le \sup B \\
\end{array}
[/tex]
I then added the two inequalities to give:
[tex]
a + b \le \sup A + \sup B
[/tex]
Since this holds true for any a and b, sup A+sup B is an upper bound on the set A+B. This is where I am a bit stuck. I can't really see why sup(A+B) isn't strictly equal to sup A + sup B. I tried coming up with a few example sets A and B but that didnt really help anything.
Any suggestions?
Thanks,
Dan,