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## Homework Statement

Let ##f## be a real-valued function with ##\operatorname{dom}(f) \subset \mathbb{R}##. Prove ##f## is continuous at ##x_0## if and only if, for every monotonic sequence ##(x_n)## in ##\operatorname{dom}(f)## converging to ##x_0##, we have ##\lim f(x_n) = f(x_0)##. Hint: Don't forget Theorem 11.4.

## Homework Equations

Theorem 11.4: Every sequence has a monotonic subsequence.

Theorem 11.8: ##(s_n)## has a sub sequential limit ##L \epsilon \mathbb{R}## and ##(s_n)## converges, then ##\lim s_n = L##.

## The Attempt at a Solution

Proof: ##\rightarrow## Suppose ##f## is continuous. Then for all sequences ##(x_n)## in ##\operatorname{dom}(f)## that converge to ##x_0##, we have that ##\lim f(x_n) = f(x_0)##. A monotonic sequence ##(a_n)## in ##\operatorname{dom}(f)## that converges to ##x_0## is a sequence in ##\operatorname{dom}(f)## that converges to ##x_0##. So for all monotonic sequences ##(a_n)## in ##\operatorname{dom}(f)## that converge to ##x_0##, we have ##\lim f(a_n) = f(x_0)## as desired.

##\leftarrow## Suppose for all monotonic sequences ##(a_n)## in ##\operatorname{dom}(f)## that converge to ##x_0##, we have ##\lim f(a_n) = f(x_0)##. Suppose ##(a_m)## is a sequence in ##\operatorname{dom}(f)## such that ##\lim a_m = x_0##. By 11.4 and 11.8, we have ##a_m## has a monotonic subsequence ##(a_{m_k})## that converges to ##x_0## and ##\lim f(a_{m_k}) = f(x_0)##. We also have the sequence ##(b_k)## defined by ##b_k = f(a_{m_k})## is a subsequence of the sequence ##(b_m)## defined by ##b_m = f(a_m)##. So ##(f(a_m))## has a sub sequential limit ##(f(x_0))##. We need to show ##(f(a_m))## converges.....

Also i'm not sure if when I want to say a function converges since a function is a set, Do i have to define a new sequence like ##(b_k)## defined as ##b_k = f(x)##, or Should I write it as ##(f(x))##?