Difference between revisions of "Dave's Recitation Sln 08/28/08 MA504Fall2008danielli" - Rhea

Latest revision as of 05:08, 29 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 $

Done.

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.

Dave's Reply: Aight, then the above proof deals with the case where $ A $ and $ B $ are bounded.

Addendum

Case Two: WLOG $ A $ unbounded $ B $ bounded. Then $ \sup A+B = \infty, \sup A = \infty, \sup B = \beta $ And using extended real number system properties $ \infty = \sup A + B = \sup A + \sup B = \infty + \beta = \infty $.

Case Three: Both unbounded. Then $ \sup A+B = \infty, \sup A = \infty, \sup B = \infty $ And using extended real number system properties $ \infty = \sup A + B = \sup A + \sup B = \infty + \infty = \infty $.

2)

Note: This needs bounded/unbounded cases stuff too.

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 $ is an upper bound of $ A $

Then $ - \beta \geq - \alpha $

Then $ \alpha \geq \beta $

$ \therefore - \alpha $ is the glb of $ -A $

Done.

3)

Let $ A \neq \varnothing \subseteq \mathbf{R}, \alpha = \sup A < \infty, \exists \delta > 0 \ni \forall a,b \in A \vert a - b \vert \geq \delta $. We will show that $ \alpha \in A $.

Assume $ \alpha \not \in A $, then $ \forall a \in A, a < \alpha $. Also given $ a \in A, \exists b \in A \ni a < b < \alpha $ else $ \forall b \in A, b \leq a $ and then $ a = \alpha $, contradiction. (1)

So, choose $ c_1 \in (a,\alpha) $, $ a + \delta \leq c_1 $ by our $ \delta $ property.

And we will show by induction that $ a + n \delta \leq c_n $. (2)

We've established the base case.

Assume $ a + n \delta \leq c_n $, we will show that $ a + (n+1) \delta \leq c_{n+1} $ for some $ c_{n+1} \in (c_n, \alpha ) $ (we have $ c_{n+1} \in (c_n, \alpha ) $ by (1) ).

By this set's property we know that $ \vert c_{n+1} - c_n \vert \geq \delta $, and since $ c_{n+1} > c_n $ we know that $ c_{n+1} \geq \delta + c_n $ and then using our inductive hypothesis this is $ \geq a + n \delta + \delta = a + (n+1) \delta $.

Which shows that $ a + (n+1) \delta \leq c_{n+1} $ for some $ c_{n+1} \in (c_n, \alpha ) $, which completes the inductive proof.

Now, by the Archimedean property and (2) $ \exists k \ni a + (k-1) \delta \leq \alpha < a + k \delta $. But this suggests that either $ \alpha $ is not the lub, or $ \alpha \in A $. Contradiction.

$ \therefore \alpha \in A $.

Done.

4)

Note: This needs to be split into cases for $ x > 0 $ and $ x \leq 0 $. I assume it's positive here. If it's negative use $ \inf A $ instead of $ \sup $, follow mutatis mutandis, and Godspeed.

Let $ A \neq \varnothing \subseteq \mathbf{R}, A $ bounded (finally), fix $ x, y \in \mathbf{R} $, and set $ T := \{ ax+y : a \in A \} $. We find $ \sup T $

By lub property we have $ \alpha = \sup A \in \mathbf{R} $.

My intuition says $ \sup T = x \alpha + y $ and we will proceed to check this.

i) We show that $ \forall \gamma \in T, \gamma \leq x \alpha + y $.

But since $ \gamma \in T, \exists \gamma ' \in A \ni x \gamma ' + y = \gamma $

And since $ \forall \gamma ' \in A, \gamma ' \leq \alpha $

$ x \gamma ' \leq x \alpha $ (This is why a split in cases should be needed)

$ x \gamma ' + y \leq x \alpha + y $

And thus we've established that it's an upper bound.

ii) We show that if $ \beta $ is a ub of $ T $, then $ x \alpha + y \leq \beta $.

$ \exists \beta ' \in \mathbf{R} \ni x \beta ' + y = \beta $, then with some algebra we establish the above claim by using the fact that $ \alpha $ is a lub.

Done.