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I did the first one by contradiction and a lot of cases. | I did the first one by contradiction and a lot of cases. | ||
| − | <math> | + | Given <math>A, B \neq \varnothing \subseteq \mathbf{R}, A+B = \{ a+b \vert a \in A, b \ in B \}</math>, and I'll let <math>\alpha = \sup A, \beta = \sup B</math> |
| − | + | WTS: <math>\sup(A+B) = \alpha + \beta</math> | |
| − | + | First of all, by the lub property of <math>\mathbf{R}</math> we know that these exist. | |
| − | + | Now we show that <math>\alpha + \beta</math> is an upper bound of <math>A+B</math> by contradiction. Thus <math>\exist a,b \in A,B \ni a+b > \alpha + \beta </math> | |
<math>(\alpha - a) + (\beta - b) < 0</math> | <math>(\alpha - a) + (\beta - b) < 0</math> | ||
| − | <math> | + | WLOG assume <math>(\alpha - a) < 0</math>, then <math>\exists a \in A \ni a > \alpha, but \alpha = \sup A</math>, contradiction. |
| − | <math>\therefore \alpha + \beta is an upper bound of A+B</math> | + | <math>\therefore \alpha + \beta</math> is an upper bound of <math>A+B</math> |
Revision as of 04:40, 28 August 2008
I did the first one by contradiction and a lot of cases.
Given $ A, B \neq \varnothing \subseteq \mathbf{R}, A+B = \{ a+b \vert a \in A, b \ in B \} $, and I'll let $ \alpha = \sup A, \beta = \sup B $
WTS: $ \sup(A+B) = \alpha + \beta $
First of all, by the lub property of $ \mathbf{R} $ we know that these exist.
Now we show that $ \alpha + \beta $ is an upper bound of $ A+B $ by contradiction. Thus $ \exist a,b \in A,B \ni a+b > \alpha + \beta $
$ (\alpha - a) + (\beta - b) < 0 $
WLOG assume $ (\alpha - a) < 0 $, then $ \exists a \in A \ni a > \alpha, but \alpha = \sup A $, contradiction.
$ \therefore \alpha + \beta $ is an upper bound of $ A+B $
