Question from my homework:
A function $f \colon S \to \mathbb{R}$ is said to be decreasing on the domain $S \subset \mathbb{R}$ if for every $x, y \in S$ with $y > x$, we have $f(y) < f(x)$. Show that if $f \colon \mathbb{R} \to \mathbb{R}$ is decreasing and bounded on $\mathbb{R}$, then the limits $\lim_{x \to \infty} f(x)$ and $\lim_{x \to -\infty} f(x)$ both exist.
Caution: While this claim is clearly related to the theorem that bounded monotone sequences always converge, it is not an immediate corollary of it. You will have to use the definition of the limit directly.
(Original picture of the problem here.)
My proof:
Let $ S = \{f(x) | x \in \mathbb{R} \} $, since $f$ is bounded this set has both an infimum and supremum. (Is this true? I tried to prove it, but have no idea as it seems fairly obvious.) Now, I claim that $\lim_{x\to \infty} f(x) = \inf(S) = L$.
By definition, given $\epsilon > 0$, there exists $K \in \mathbb{R}$ such that $x>K \implies |f(x) - L| < \epsilon$.
Now since $f(x) \geq L$ by definition of the infimum, choose $x_1$ such that $f(x_1) < L + \epsilon$ for $x_1 > K$. But since $f$ is a decreasing function we have $f(x) \leq f(x_1)$, and thus $$ L - \epsilon < L \leq f(x) \leq f(x_1) < L + \epsilon $$ which proves our claim.
I use a similar argument for the supremum. Could anyone help me out on making this more rigorous as I feel as though I've skipped some steps.
I think you are almost there. I would proceed like this: For $\epsilon > 0$ be a given, there exists an $x^{*}$ such that $f(x^{*}) < L + \epsilon $. So let $K = |x^{*}| + 1 > 0$, then for all $x$'s such that $x > K$ we have: $f(x) < f(K) < f(x^{*}) < L + \epsilon$. But $L < f(x)$ since it is $\text{inf}\{f(x): x \in \mathbb{R}\}$. So $L - \epsilon < L < f(x) < L + \epsilon$. This means $|f(x) - L| < \epsilon$ and we're done.