I need a help. Can somebody check,do I understand correctly a proof of next exercise.
Let $I \subseteq \mathbb R$ be an interval and $f:I \to \mathbb R$ be increasing function on $I$. Suppose $c \in I$ is not an endpoint of $I$. Show that $f$ is continuous at $c$ if and only if there exists a sequence $(x_n)$ in $I$ such that $(x_n)<c$ for $n=1,3,5…$ and $(x_n)>c$ for $n=2,4,6$; and such that $c=\lim(x_n)$ and $f(c)=\lim f(x_n))$.
$j_f(c)$ is jump of $f$ at point $c$.
$\textbf{Proof in the book}$:
$"\Rightarrow"$ is trivial.
$"\Leftarrow"$
Since $0\leq j_f(c)\leq f(x_{2n})-f(x_{2n+1})$ $\textbf{it follows that $j_f(c)=0$}$. So $f$ is continuous at $c$. The question is, why does it follow?
$\textbf{My assumption}$: I think, here was used the Squeeze Theorem. If $0\leq j_f(c) \leq f(x_{2n})-f(x_{2n+1})$ then it has to be $0\leq j_f(c) \leq 0$. Now the question is, why is it $f(x_{2n})-f(x_{2n+1})=0$. I think because $(x_n)$ converges to $c$ and thus every subsequence converges to $c$ so $f(x_{2n})-f(x_{2n+1})=f(c)-f(c)=0$. Am I right? Can somebody explain me this,please. Thank you for help!
Best Answer
Since $f$ is increasing and $x_{2n+1} < c < x_{2n}$ we have $$f(x_{2n+1}) \le \lim_{x \to c^-} f(x) \le \lim_{x \to c^+} f(x) \le f(x_{2n})$$ and thus $$j_f(c) := \lim_{x \to c^+} f(x) - \lim_{x \to c^-} f(x) \le f(x_{2n}) - f(x_{2n+1}).$$
It is not true that $f(x_{2n}) \le f(x_{2n+1})$ for a given $n$. But it is true in the limit as $n \to \infty$.
Since $0 \le j_f(c) \le f(x_{2n}) - f(x_{2n+1})$ holds for each $n$, we can take the limit as $n \to \infty$ of the right-hand side to get $0 \le j_f(c) \le 0$ as you mentioned in your post.