I have been stuck on this problem for a while and can't form a good argument for what I think is right. The statement is:
"Let $f$ be a function such that $(-\infty, a] \subseteq \text{dom}(f)$ for a given $a \in \mathbb{R}$ (that is, $f$ is defined for all $x \leq a$). If $f$ is monotonic decreasing and bounded above then $\lim_{x \xrightarrow{} -\infty}{f(x)} = \sup_{x \in \text{dom}(f)}{f(x)}$."
First, I want to know if the statement is true. Intuitively, it makes sense, but I might have missed some crazy edge case of a function.
Here is my incomplete proof:
"Let $M = \sup_{x \in \text{dom}(f)}{f(x)}$. Since $M$ is the supremum, then there exists a $y \in \text{range}(f)$ such that $M - \epsilon < y \leq M$ for all $\epsilon > 0$. Since $y$ is an element of the range of the function, then there exists an $x_0 \in \text{dom}(f)$ such that $y = f(x_0)$. We can then write $M - \epsilon < f(x_0) \leq M$ for all $\epsilon > 0$. Since $f$ is decreasing we know that $f(x) \leq f(x_0)$ for all $x \in \text{dom}(f)$ such that $x > x_0$. We also know that because $f$ is bounded above that $f(x) \leq M$ for all $x > x_0$."
Now, to complete the proof, I'd have to argue that $M - \epsilon < f(x)$ for all $x > x_0$ and $\epsilon > 0$, and then some. But what would would that even look like? Am I missing some corollary, or is the statement itself bogus?
Your statement is correct, and your proof is not too far off.
To have it mentioned: The inequality $M-ε < y \leq M$ does not hold for all $ε>0$; it holds that for each $ε>0$ you find a $y$ which satisfies the inequality.
Now to prove your statement, let $L = \lim_{x\to -\infty} f(x)$ and $M = \sup_{x\in \operatorname{dom}(f)} f(x)$. By definition of supremum, for every $ε>0$ we find $x_0\in \operatorname{dom}(f)$ with $M-ε < f(x_0) \leq M$. Moreover, for any $x\leq x_0$, we have that $f(x)\geq f(x_0)$ (since $f$ is decreasing), so $$M-ε < f(x_0) \leq f(x) \leq M < M+ε \text{ for all } x\leq x_0$$ Rewriting this implies precisely that for all $ε>0$, there exists $x_0\in \operatorname{dom}(f)$ such that $$|f(x) - M| < ε \text{ for all } x\leq x_0,$$ which is precisely the definition of a limit as $x\to-\infty$, so $M = L$.