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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 A, B \neq \varnothing \subseteq \mathbf{R}, A+B = \{ a+b \vert a \in A, b \ in B \} | + | <math>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</math> |
| − | WTS: \sup(A+B) = \ | + | <math>WTS: \sup(A+B) = \alpha + \beta</math> |
| + | |||
| + | <math>First of all, by the lub property of \mathbf{R} we know that these exist.</math> | ||
| + | |||
| + | <math>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 </math> | ||
| + | |||
| + | <math>(\alpha - a) + (\beta - b) < 0</math> | ||
| + | |||
| + | <math>WLOG assume (\alpha - a) < 0, then \exists a \in A \ni a > \alpha, but \alpha = \sup A, contradiction.</math> | ||
| + | |||
| + | <math>\therefore \alpha + \beta is an upper bound of A+B</math> | ||
Revision as of 04:39, 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 $
