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.

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.

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood