Convergence of a continuous function related to a monotonic sequence

[fishturtle1]

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 $(f(x))$?

 

[andrewkirk]

For the second direction, picking a sequence is not going to help, as showing it converges will not lead anywhere.
Instead, assume $f$ is not continuous and use that to construct a sequence $(x_n)$ that converges to $x_0$ such that $f(x_n)$ does not converge to $f(x_0)$. That will prove what we want by contradiction.

It is easy to do using the axiom of choice. Start by observing that if $f$ is not continuous at $x_0$ then there exists some $\epsilon>0$ such that for any $\delta$, there exists $x\in(x_0-\delta,x_0+\delta)$ such that $|f(x)-f(x_0)|>\epsilon$. Think of a way to get a sequence of progressively smaller deltas.

Doing it without the axiom of choice may be more difficult (I'm not even sure if it's possible). But the question doesn't say the axiom may not be used.

 

[fishturtle1]

For the second direction, picking a sequence is not going to help, as showing it converges will not lead anywhere.
Instead, assume $f$ is not continuous and use that to construct a sequence $(x_n)$ that converges to $x_0$ such that $f(x_n)$ does not converge to $f(x_0)$. That will prove what we want by contradiction.

It is easy to do using the axiom of choice. Start by observing that if $f$ is not continuous at $x_0$ then there exists some $\epsilon>0$ such that for any $\delta$, there exists $x\in(x_0-\delta,x_0+\delta)$ such that $|f(x)-f(x_0)|>\epsilon$. Think of a way to get a sequence of progressively smaller deltas.

Doing it without the axiom of choice may be more difficult (I'm not even sure if it's possible). But the question doesn't say the axiom may not be used.

i'm not sure how to use the axiom of choice here. The wikipedia entry says for any set X of nonempty sets, there exists a choice function f defined on X. But that doesn't seem applicable because we don't have a set of nonempty sets?

Going off post #2:
if $f$ is not continuous at $x_0$ then there exists some $\epsilon>0$ such that for any $\delta$, there exists $x\in(x_0-\delta,x_0+\delta)$ such that $|f(x)-f(x_0)|>\epsilon$. By 11.4, there is some monotonic subsequence $(x_{n_k})$ of $(x_n)$. By 11.8, $(x_{n_k})$ converges to $x_0$. I think that $x_{n_k}$ being monotonic gives us smaller deltas because we are getting closer and closer to the limit but I'm not sure how to put this in words.

 

[andrewkirk]

For simplicity of notation, use the standard 'ball' notation that $B_r(x)$ denotes the 'open ball' of radius $r$ centred at $x$ which, for $\mathbb R$ means that $B_\delta(x)$ is the interval $(x-\delta,x+\delta)$. Now the sequence $(\delta_n)$ such that $\delta_n=1/n$ is monotonically reducing to 0.

Also, the collection

$$C = \{B_{\delta_n}(x_0)\ :\ \delta_n=1/n\}$$

is a collection of nonempty sets. Define $A_n$ to be the set of points $x'$ in $B_{\delta_n}(x_0)$ such that $|f(x')-f(x_0)|>\epsilon$. Do we know whether any given set $A_n$ is empty or not, given what we assumed about $f$ not being continuous? If we were able to conclude that every $A_n$ is nonempty, can we say anything about whether there exists a sequence $(x_n)$ such that $x_n\in A_n$ (think axioms)? To what would that sequence converge? Would $(f(x_n))$ converge to $f(x)$? Can we get a contradiction?