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The following question is extracted from Royden's Real Anlysis $4$th edition, question $36$ at page $53$:

Let $f$ be an increasing function on the open interval $I$. For $x_0 \in I,$ show that $f$ is continuous at $x_0$ if and only if there are sequences $\{ a_n \}$ and $\{ b_n \}$ in $I$ such that for all $n$, $a_n < x_0 < b_n$ and $\lim_{n \rightarrow \infty}{[f(b_n) - f(a_n)]} = 0.$

I am trying to prove the $(\Rightarrow)$ direction.

I have been trying to do the following:

Since $f$ is continuous at $x_0$, for any $n \in \mathbb{N}$, there exists $\delta_n >0$ such that for all $x$, $|x - x_0| < \delta_n \Rightarrow |f(x) - f(x_0)| < \frac{1}{n}.$

However, I fail to see that why must there $a_n < x_0 < b_n .$ In the first place, how do we know that $a_n$ and $b_n$ are different from $x_0$?

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Instead of directly using the definition of continuity, use the connection between continuous functions and converging sequences given by Proposition 21 of Section 1.6 in your text.

Because $I$ is open, there is an $r > 0$ for which the interval $(x_0 - r,x_0 + r)$ is contained in $I$. Choose a natural number $N$ such that $1/N < r$. Then the sequences defined by $a_n = x_0 - 1/(N + n)$, and $b_n = x_0 + 1/(N + n)$ are in $I$ such that for each $n$, $a_n < x_0 < b_n$. Moreover, $\{a_n\}$ and $\{b_n\}$ converge to $x_0$ so that by Proposition 21 of Section 1.6, $\lim_{n\to\infty} f(a_n) = f(x_0) = \lim_{n\to\infty} f(b_n)$. Thus, by Theorem 18 of Section 1.5, $\lim_{n\to\infty} (f(b_n) - f(a_n)) = f(x_0) - f(x_0) = 0$.

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Hint: Start with $a_n = x_0 - \frac{1}{n}$ and $b_n = x_0 + \frac{1}{n}$. Then use the fact that $I$ is open to show that at some point both sequences are in $I$. Conclude using continuity at $x_0$.

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