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'''Bobby's criticism:''' The sets A and B are not bounded above so you cannot apply the LUB property! If they are, your proof looks good. If not, your proof may have problems manipulating <math>\infty=\sup A</math>. Break it into cases. | '''Bobby's criticism:''' The sets A and B are not bounded above so you cannot apply the LUB property! If they are, your proof looks good. If not, your proof may have problems manipulating <math>\infty=\sup A</math>. Break it into cases. | ||
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| + | 2) | ||
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| + | Let <math>A \neq \varnothing \subseteq \mathbf{R}, -A := \{ -a : a \in A \}</math>, we will show that <math>- \sup A = \inf (-A)</math>. Let <math>\alpha = \sup A</math> (exists by lub property of <math>\mathbf{R}</math>) | ||
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| + | We first show that <math>- \alpha</math> is a lower bound. I.e., we WTS <math>\forall \gamma \in -A, - \alpha < \gamma</math> | ||
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| + | But this is true <math>\because \alpha > - \gamma, \forall - \gamma \in A</math> by <math>\alpha</math> being an upper bound of <math>A</math>. | ||
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| + | Now we show that this is the greatest lower bound. Let <math>\beta</math> be a lower bound of <math>-A</math> | ||
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| + | Then <math>- \beta<math> is an upper bound of <math>A</math> | ||
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| + | Then <math>- \beta \geq - \alpha</math> | ||
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| + | Then <math>\alpha \geq \beta</math> | ||
| + | |||
| + | <math>\therefore - \alpha</math> is the glb of <math>-A</math> | ||
| + | |||
| + | <math>\halmos</math> | ||
Revision as of 05:59, 28 August 2008
- 1
I did the first one by contradiction. I noted that the first time I looked through it I forgot to verify their existance via the lub property of the reals, and I still wonder whether the proof needs cases by whether the supremum is infinite or finite.
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 $
Now we show that it is the least upper bound. Let $ \delta $ be an upper bound of $ A+B $, we need to show that $ \delta \geq \alpha + \beta $.
Assume not, then $ \delta < \alpha + \beta $, let $ c = \alpha + \beta - \delta > 0 $ and note $ \delta = \alpha + (\beta - c) $. But since $ \delta $ is an upper bound of $ A+B $ then $ \forall a,b \in A,B, a+b \leq \alpha + b \leq \delta = \alpha + \beta - c $. But then $ \forall b \in B, b \leq \beta - c $ so we have an upper bound for $ B $ that's smaller than the lub of $ B $, contradiction.
Therefore, $ \alpha + \beta $ is the lub of $ A+B $
Bobby's criticism: The sets A and B are not bounded above so you cannot apply the LUB property! If they are, your proof looks good. If not, your proof may have problems manipulating $ \infty=\sup A $. Break it into cases.
2)
Let $ A \neq \varnothing \subseteq \mathbf{R}, -A := \{ -a : a \in A \} $, we will show that $ - \sup A = \inf (-A) $. Let $ \alpha = \sup A $ (exists by lub property of $ \mathbf{R} $)
We first show that $ - \alpha $ is a lower bound. I.e., we WTS $ \forall \gamma \in -A, - \alpha < \gamma $
But this is true $ \because \alpha > - \gamma, \forall - \gamma \in A $ by $ \alpha $ being an upper bound of $ A $.
Now we show that this is the greatest lower bound. Let $ \beta $ be a lower bound of $ -A $
Then $ - \beta<math> is an upper bound of <math>A $
Then $ - \beta \geq - \alpha $
Then $ \alpha \geq \beta $
$ \therefore - \alpha $ is the glb of $ -A $
$ \halmos $
