I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Show that $\sup(−x_n) = −\inf (x_n)$ for any sequence $(x_n)$ in $\mathbb{R}$. Conclude that $\lim$ $\sup(−x_n) = −\lim$ $\inf (x_n)$.
My Proof:
Part One:
Let $B$ be the set of upper bounds to the sequence $(-x_n)$, and let $z$ be the supremum such that $ \forall b \in B(z\le{b})$. Since $B$ is the set of upper bounds, for any $b \in B$:
$\forall n \in N(-x_n\le{b})$
$\forall n \in N(x_n\ge{-b})$
Which means that $-b$ is a lower bound of $(x_n)$, so if we define $-B = \{-b \vert b\in B\}$, $-B \subseteq L$, where $L$ is the set of lower bounds of $(x_n)$. Suppose $a\in L$, then $\forall n\in N(x_n\ge{a})$, so $\forall n\in N(-x_n\le{-a})$ so $-a$ is an upper bound of $(-x_n)$, so $-a\in B$ and therefore $a\in -B$, so $L\subseteq -B$. Thus, $-B=L$ and $-B$ is the set of all lower bounds of $(x_n)$. $z$ is the supremum, so $z \in B$, and since $\forall b \in B(z\le{b})$, $\forall b \in B(-z\ge{-b})$, so $\forall b'\in-B(-z\ge{b'})$ and so $-z$ is the infimum of $(x_n)$. Thus,
$\sup(-x_n) = z = - (-z) = -\inf(x_n)$
Part Two:
$\lim$ $\sup(-x_n)$ =
$\inf_m$ $\sup_{n\ge{m}} (-x_n)$ =
$\inf_m$ $(-\inf_{n\ge{m}} (x_n))$ =
*Notice that $(-\inf_{n\ge{m}} (x_n)$ is a sequence, so define $(x_m)$ = $(\inf_{n\ge{m}} (x_n))$
$\inf$ $(-x_m)$ =
$-\sup$ $(x_m)$ =
$-\sup_m$ $(\inf_{n\ge{m}} (x_n))$ =
$-\lim$ $\inf(x_n)$
Thus, the statement we proved in Part One shows us that $\lim$ $\sup(-x_n)$ = $-\lim$ $\inf(x_n)$