One-sided limits functions and monotony(proposition)

Question

This is the proposition: A monotone function $f:D\rightarrow\mathbb R$, where $D$ is included in $\mathbb R$, has one-sided limits in every accumulation point of $D$ (not necessarily in $D$.

I actually have the proof, but I don't like because it uses an analogy: it uses a arbitrary sequence $x_n$ that tends to an arbitrary accumulation point of $D$ (from one side) to prove that $f(x_n)$ tends to a finite or infinite limit.

Anyway, as the sequence doesn't have to be ordered, it doesn't have to be monotone. But the proof assumes that they are ordered because "is the same thing", in my context, it means that no matter the order, the sequence still has a limit. I don't understand why it is the same thing, what does make it an analogy? What's the key? Thanks in advance!

Answer 1

Hint:

Well, can you prove that every sequence $\left(x_n\right)_{n\in\mathbb{N}}\subseteq\mathbb{R}$ has a monotone subsequence $\left(x_{k_n}\right)_{n\in\mathbb{N}}$? Give it a try and infrom us about your resutls.

Edit: Now, the sequence $(x_n)$ you are given converges to an accumulation point $x$ of $D$ and $x_n\leq x$ - from "one side". Now, there is a subsequence $(x_{k_n})$ of $(x_n)$ which is monotone. Since it is a subsequence of $(x_n)$ and $x_n\to x$, we also have that: $$x_{k_n}\to x$$ Now, let $y_n=x_{k_n}$ and you have a sequence that is monotone, converges to $x$ and $y_n\leq x$. This is why it does not matter to consider your sequence monotone.

Further Edit: Due to $(y_n)$ being monotone, we also have that $(f(y_n))$ is monotone and, hence, it is either convergent or diverges to $\pm\infty$ (assume the case where $f$ is increasing, so it should be $+\infty$). Let now $l\in\mathbb{R}\cup\{+\infty\}$ be that limit, so $$\lim_{n\to\infty}f(y_n)=l$$

1. If $l=+\infty$, then, evidently $f(x_n)\to+\infty$, since $f$ is monotone. However, for reasons of completeness, let us prove it. Let $M>0$. Then, there exists a $n_0\in\mathbb{N}$ such that, for every $n\geq n_0$: $$f(x_{k_n})=f(y_n)>M$$ Note now that, since $x_n\to x$ and $x_n\leq x$ we have that there exists a $n_1\in\mathbb{N}$ such that, for every $n\geq n_1$: $$|x_n-x|<\epsilon=x-y_{n_0}>0$$ So, it comes that we have: $$x_n\geq y_{n_0}=x_{k_{n_0}}$$ And, due to $f$'s monotonicity, we have that: $$f(x_n)\geq f(y_{n_0})$$ So, let $N=\max\{n_0,n_1\}$ (actually, $N=n_1$). Then for every $n\geq N$ we have that: $$f(x_n)\geq f(y_{n_0})>M$$ and the proof is complete.
2. If $l\in\mathbb{R}$, then again $f(x_n)\to l$. So, let $\epsilon>0$. Then we have that there exists a $n_0\in\mathbb{N}$ such that, for every $n\geq n_0$: $$|f(x_{k_n})-l|=|f(y_n)-l|<\epsilon$$ Now, note the $f$'s and $(y_n)$'s monotonicity imply that: $$0<l-f(y_n)<\epsilon$$ (Actually, the first inequality may be $\leq$, but then the requested is trivial). Let $$\epsilon'=x-y_{n_0}=x-x_{k_{n_0}}$$ Then, there exists $n_1\in\mathbb{N}$ such that, for every $n\geq n_1$: $$|x_n-x|<\epsilon'$$ which implies that: $$x_n\geq x_{k_{n_0}}$$ So, for $N=\max\{n_0,n_1\}$ we have that, for every $n\geq N$: $$l-f(x_n)<\epsilon$$ and the proof is complete.