$$f: Df \rightarrow R, Df \subset R$$
f is monotone function and $Rf$ is interval.
Then $f$ is continuous.
Proof:
Suppose that $$\exists \epsilon >0 : \forall \delta >0 \, \, \exists x\in (Df \cap |x|<x_o+\delta) : |f(x)-f(x_o)| \ge \epsilon$$
$$\delta - fixed; \delta - \frac{1}{n}, n\in N$$
$$\exists x_n : (|x_n|< x_0 + \frac{1}{n} \, \, \land x_n\in Df) : |f(x)-f(x_o)\ge \epsilon $$
$$(x_n)_{n\in N} \, \, x_o-\frac{1}{n}< x_n < x_0 +\frac{1}{n}$$
$$\forall_{n\in N} x_n \in (Df \land \lim_{n \to \infty}x_n=x_o)$$
First of all, I wanted to understand whether the fact that $\exists x: |x|<x_o+\delta $ is derived from the fact that $Rf$ is interval and every monotone function that is bounded has a limit or is it something else?
$$D^-f = \lbrace x \in Df | x<x_o \rbrace$$
$$D^+f=\lbrace x \in Df | x>x_o \rbrace$$
$$\forall_{n\in N} x_n \in D^-_f \cup D^+f$$
In one of the sets there is infinite amount of sequence members ( I've heard that if $x_o$ is a limit then on it's neighbourhood there are infinite amount of members, but in this instance it not about neighbourhood of $x_o$, is it?).
Supposie that $D^-f $ has infinite members of $(x_n)_{n\in N}$. $(x_{n_k})_{k \in n}$ is a subsequence of $(x_n)_{n \in N}$. $\forall k \in N \, \, x_{n_k} \in D^-_f$. $\lim_{k \to \infty}x_{n_k}=x_o$.
Suppose that f is increasing. Then $\forall k \in N \, \, f(x_{n_k})\le f(x_o). \, \, \, \forall k \in N \, |f(x_{n_k})-f(x_o)| \ge \epsilon$
I don't understand the last expression, at the beginning of theorem it was supposed that exists such Epsilon where for any Delta the absolute value between $f(x)$ and $f(x_o)$ is bigger or equal than Epsilon. But here the subsequence simply belongs to $D_f$ and it's not clear that for all $k$ such equality will be valid? What did I miss here?
$$\forall k \in N \, \, f(x_{n_k}) \le f(x_o) - \epsilon$$
$$\forall x\in Df \, \, k<x_o \, \Rightarrow f(x) \le f(x_o) - \epsilon$$
$$\lim_{k \to \infty}x_{n_k} - \exists k_o \, \, \forall k \in N \Rightarrow k \ge k_o \Rightarrow |x_{n_k}| \le x_o + \epsilon $$
I don't understand the meaning behind those last few expressions, where it is showed that f is continuous?