Assumgin A,B, two set are upper-bounded. I need to prove that A Union B is also upper bounded and the supremum is max(supA, supB).
This question can be explained intuitivly, but how do you prove it in a formal/mathemtical way?
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Nick Peterson
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captain dragon
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3How are you defining the "union" of two sequences? – Nick Peterson Oct 25 '13 at 15:47
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1sorry, I ment "set", not "sequnces" – captain dragon Oct 25 '13 at 15:49
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Hint:
Step one is to show that the set $A\cup B$ is bounded above. We know that $A$ is bounded above by $\sup A$, and that $B$ is bounded above by $\sup B$. We also know that $\sup A\leq\max\{\sup A, \sup B\}$ and $\sup B\leq\max\{\sup A,\sup B\}$, by definition of the maximum. How can you combine these to show that $\max\{\sup A,\sup B\}$ is an upper bound on $A\cup B$?
Step two: having shown that $\max\{\sup A,\sup B\}$ is an upper bound, you must show that it is the least upper bound. So, suppose that you have an upper bound $M$ on $A\cup B$ which satisfies $M<\max\{\sup A,\sup B\}$. Why is this a contradiction if $M<\sup A$? What about if $M<\sup B$? Show these are both impossible, and you're done -- since being less than the maximum require that one or the other holds.
Nick Peterson
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These hints helped me greatly with a homework problem. Thank you very much! – Valentino Sep 15 '14 at 16:11