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.
