HW 1

Proof:

1. $\Rightarrow$
   1. Assume $|a-b|\le c$ for some positive $c$
   2. then $-c\le a-b\le c$. Adding $b$ to each side we get $b-c\le a\le b+c$
2. $\Leftarrow$
   1. Assume $b-c\le a\le b+c$
   2. Then, subtracting $b$ from both sides we get $-c\le a-b\le c$
   3. thus $|a-b|\le c$.

Proof

1. As both the previous sets are bounded, let $M_s,M_t$ be $\sup s$ and $\sup t$ respectively. Then, w.l.o.g assume $M_s\ge M_t$. With this, it stands that $M_s$ will also be an upper bound of $S\cup T$ as it is larger than any element in that set. Taking any element $x\in \sup S\cup T$, either $x\in S$ or $x\in T$. Then, if $x\in S$ we have it so $x\le M_s$. If $x\in T$, we have it so $x\le M_t\le M_s$, thus it is bounded by $M_s$. Use the same logic for infimum.
2. W.L.O.G assume $\inf S\le \inf T$. Then, by definition it is smaller than any element in $S$. It is also smaller than any element in $T$ because for any $t\in T$ we have it so $\inf S\le \inf T\le t$. Thus, $\min \{\inf S,\inf T\}=\inf S$ is a lower bound for $S\cup T$. Now assume there was another lower bound $L$ of $S\cup T$. Then $\inf S\ge L$ because its the greatest lower bound of $S$ and $\inf S\le \inf T$. By definition it is the greatest lower bound of $S$, and if there was a greater bound $L$ then either $\inf S\le L\le \inf T$, which means there are such elements in $S/T\subset S\cup T$ which are less than $L$ serving as a contradiction, or $\inf S\le \inf T\le L$ which means there are elements in both $S$ and $T$ that are less than $L$.

Proof:

1. Firstly, notice that $\sqrt{2}$ is an upper bound, by definition. Now, for the sake of contradiction, assume there is another upper bound $M<\sqrt{2}$. Then, by the density of $Q$, since both $\sqrt{2}$ and $M$ are real numbers, there exists a rational between $\sqrt{2}$ and $M$ serving as a contradiction for $M$ being an upper bound.

Proof: todo hint: show the other expanded form, include triangle inequality with something like $a=|b+(a-b)|$