Monotonic sequence definition of Continuity of a function

[Hall]

Homework Statement

Main body

Relevant Equations

Main body.

Question: There is a function $f$, it is given that for every monotonic sequence $(x_n) \to x_0$, where $x_n, x_0 \in dom(f)$, implies $f(x_n) \to f(x_0)$. Prove that $f$ is continuous at $x_0$

Proof: Assume that $f$ is discontinuous at $x_0$. That means for any sequence $(x_n)$, in $dom(f)$, converging to $x_0$ we don't have $\lim f(x_n) = f(x_0)$, symbolicallyFor every N there exists, at least one, $k > N$ such that
$$
|x_k - x_0| < \delta ~ \text{but} ~ |f(x_k) f(x_0)| \geq \varepsilon
$$
Doubt 1: I haven't said anything about $\delta$ and $\varepsilon$ before, and simply introduced them to mean the difference is small,is this how we do it formally?
Now, as for every $N$ we have "a" $k$ satisfying those properties, therefore, we have all those $k$ forming a subset of $\mathcal{N}$ like thus
$$
k_1 > N_1 ~~\text{satisfying those properrties}$$
$$k_2 > N_2 > k_1 ~~ \text{satisfying those properties}$$
$\vdots$
$$k_i > N_i > k_{i-1}$$
$\vdots$

and hence, $(x_k)$ is a sub-sequence of $(x_n)$, and
$$\lim (x_k) \to x_0$$

$\since$ for every $x_k$ we have $|f(x_k) - f(x_0)| \geq \varepsilon$, but $(x_k)$ is a monotone sequence and we must have $|f(x_k) - f(x_0)| < \varepsilon$

This contradiction has arisen because our assumption that $f$ was discontinuous was wrong.

Doubt #2: Is what I have done correct?

 

[haruspex]

Assume that $f$ is discontinuous at $x_0$. That means for any sequence $(x_n)$, in $dom(f)$, converging to $x_0$ we don't have $\lim f(x_n) = f(x_0)$,

No, it means that for some sequence $(x_n)$, in $dom(f)$, converging to $x_0$ we don't have $\lim f(x_n) = f(x_0)$.

 

[Hall]

No, it means that for some sequence $(x_n)$, in $dom(f)$, converging to $x_0$ we don't have $\lim f(x_n) = f(x_0)$.

Yes, I can see it.

 

[Hall]

It doesn't bring down my proof, right? Or am I missing something else?

 

[Office_Shredder]

The proof is just bad. You're like 50% of the way there conceptually but the terrible notation doesn't string together in a way that gets you close.

The biggest issue: why is thy sequence you constructed at the end monotone?

You should start with this, using the sequence definition of convergence. If it's not clear why what I'm writing below is true, start by proving it (it's equivalent to saying $f$ is continuous when for all sequences $x_n\to x_0$ we have $f(x_n)\to f(x_0)$)

if $f$ it's discontinuous at $x_0$ then there exists $\epsilon>0$ and a sequence $x_n\to x_0$ such that for all $N>0$, there exists $k>N$ such that $|f(x_k)-f(x_0)| >\epsilon$

 

[Hall]

The biggest issue: why is thy sequence you constructed at the end monotone?

I guess you mean $k_1 \lt k_2 \lt k_3 \lt \cdots k_i \cdots \lt$ doesn't imply
$$
x_{k_1} \lt x_{k_2} \lt x_{k_3} \cdots
$$
I agree with you.

 

[Hall]

Let me mend the proof (Someone there is that doesn't love the proof):

Assume that $f$ is discontinuous at $x_0$. That means for any sequence $(x_n)$, in $dom(f)$, converging to $x_0$ we don't have $\lim f(x_n) = f(x_0)$, symbolicallyFor every N there exists, at least one, $k > N$ such that
$$
|x_k - x_0| < \delta ~ \text{but} ~ |f(x_k) f(x_0)| \geq \varepsilon
$$
As for each $N$ we have a $k$ such that $x_k$ follows those two things, we have a whole sequence of those points as $(x_k)$ (in actuality a subsequence of $(x_n)$).

As from any sequence we can extract a monotone sequence, let's say $(x_{k_i})$ is a monotone sub-sequence of $(x_k)$. We have, $x_{k_i} \to x_0$, but $|f (x_{k_i} ) -f(x_0) | \geq \varepsilon$. That is, though $(x_{k_i}$ is a monotone sequence and converges to $x_0$ but $\lim f(x_{k_i}) \neq f(x_0)$. This is in pure contradiction with the given statement.

Thus, the given statement implies continuity of $f$.

 

[FactChecker]

Proof: Assume that $f$ is discontinuous at $x_0$. That means for any sequence $(x_n)$, in $dom(f)$, converging to $x_0$ we don't have $\lim f(x_n) = f(x_0)$, symbolicallyFor every N there exists, at least one, $k > N$ such that
$$
|x_k - x_0| < \delta ~ \text{but} ~ |f(x_k) f(x_0)| \geq \varepsilon
$$
Doubt 1: I haven't said anything about $\delta$ and $\varepsilon$ before, and simply introduced them to mean the difference is small,is this how we do it formally?

No. You need to introduce and quantify the $\epsilon$ and $\delta$ in some way. In this case, I think you are using them in a proof by contradiction, so you are assuming the function is discontinuous. You should say something like, "There exists an $\epsilon \gt 0$ such that $\forall \delta \gt 0$ and for each N ..."

 

[Office_Shredder]

The way you know $|x_k-x_0|$ is small is because $k>N$, and $N$ is arbitrarily big. There's no independent picking $\delta$ here.

 

[FactChecker]

You have to also change "for any sequence $(x_n)$" to "there exists a sequence $(x_n)$", and then describe the properties of that sequence consistent with your assumption that the function is discontinuous at $x_0$.

 

[Hall]

I have got another doubt which I think doesn't demand a new thread.

Problem statement:
Let f be a real-valued function whose domain is a subset of R. Show f is continuous at x_0 in dom(f) if and only if, for every sequence (x_n) in dom(f) \ {x0} converging to x0, we have lim f(x_n) = f(x0).

Solution: I'm a little confused here as of what difference $dom (f) - \{x_0\}$ will make because we always put $0 \lt |x_n - x_0| \lt \delta$, thus, in itself signifying that $x_n$ never gets $x_0$ in a $\delta-$ neighbourhood, yes, but, the possibility that it becomes equal to $x_0$ outside $\delta$ is still there, but why do care? We're only interested in $\delta-$ neighbourhood.

However, here is the proof:

(Forward implication): Given that $f$ is continuous at $x_0$, we have, by definition:
"for each $epsilon \gt 0$ there exists a $\delta \gt 0$ such that
$$
0 \lt |x - x_0| \lt \delta \implies |f(x) - f(x_0)| \lt \epsilon"
$$
Consider a sequence $(x_n)$ converging to $x_0$, with the property that
$$
x _n \neq x_0 ~~~~~~~~\textrm{for all n}
$$
For every $\delta \gt 0$, we have an $N$ such that
$$
n \gt N \implies |x_n - x_0| \delta \implies |f(x_n) - f(x_0)| \lt \epsilon
$$
Thus, $\lim f(x_n) = f(x_0)$.

(backward implication): Given that for every sequence $(x_n) \to x_0$ with the property that $x_n \neq x_0$ for all n, implies $\lim f(x_n) = f(x_0)$.

Assume, that $f$ is discontinuous at $x_0$, that is in each $\delta-$ neighbourhood of $x_0$ there is an $x$ such that
$$
\begin{equation*}
| f(x) - f(x_0) | \geq \varepsilon
\end{equation*}
$$
So, for a particular sequence $x_n = x_0 - \frac{1}{n}$ (and it satisfies both the qualifications of our given statement) we have for a sufficiently large $N$
$$
\begin{align*}
n \gt N \implies 0 \lt |x_n - x_0| \lt \delta \\
\textrm{but} ~~~ |f(x_n) - f(x_0)| \geq \varepsilon \\
\end{align*}
$$
for some $x$, thus, $\lim (x_n) = x_0$ but $\lim f(x_n) \neq f(x_0)$. But this contradicts our given statement. Thus, $f$ is continuous at $x_0$.

I mean nowhere I could see any importance of "$(x_n)~\textrm{ in}~ dom(f)-\{x_0\}$".

 

[FactChecker]

A general comment I have is that your understanding is ok, but you are not correctly using the official definitions. You need to be more careful about that.

I have got another doubt which I think doesn't demand a new thread.

Problem statement:
Let f be a real-valued function whose domain is a subset of R. Show f is continuous at x_0 in dom(f) if and only if, for every sequence (x_n) in dom(f) \ {x0} converging to x0, we have lim f(x_n) = f(x0).

Solution: I'm a little confused here as of what difference $dom (f) - \{x_0\}$ will make because we always put $0 \lt |x_n - x_0| \lt \delta$,

Is $0 \lt$ in your official definition? It is not enough to say "we always put". You have to use your official definition. Of course, an infinite sequence of $x_0$s do "converge" to $x_0$ but that sequence is not helpful in defining continuity.

(Forward implication): Given that $f$ is continuous at $x_0$, we have, by definition:
"for each $epsilon \gt 0$ there exists a $\delta \gt 0$ such that
$$
0 \lt |x - x_0| \lt \delta \implies |f(x) - f(x_0)| \lt \epsilon"
$$
Consider a sequence $(x_n)$ converging to $x_0$, with the property that
$$
x _n \neq x_0 ~~~~~~~~\textrm{for all n}
$$
For every $\delta \gt 0$, we have an $N$ such that

No. You have to use your official definition. That is in terms of "there exists a $\delta$", not $N$. In this step of your proof, you have to use the existence of $\delta$ to show that $N$ exists.

(backward implication): Given that for every sequence $(x_n) \to x_0$ with the property that $x_n \neq x_0$ for all n, implies $\lim f(x_n) = f(x_0)$.

Assume, that $f$ is discontinuous at $x_0$, that is in each $\delta-$ neighbourhood of $x_0$ there is an $x$ such that
$$
\begin{equation*}
| f(x) - f(x_0) | \geq \varepsilon
\end{equation*}
$$

Where did this $\epsilon$ come from? You have said nothing about it, but now it suddenly appears. What property does it have?

 

[Hall]

Yes, the official definition doesn't say $0\lt |x-x_0| \lt \delta$

Let $f$ be a real-valued function whose domain is a subset of R.Then $f$ is continuous at $x_0$ in dom(f) if and only if for each $\epsilon \gt 0$ there exists δ>0 such that x ∈dom(f) and $$
|x−x_0|<δ \implies |f(x)−f(x_0)| < \epsilon
$$I will, I think, benefit more if I shall be given a model proof.

 

[FactChecker]

Yes, the official definition doesn't say $0\lt |x-x_0| \lt \delta$

Let $f$ be a real-valued function whose domain is a subset of R.Then $f$ is continuous at $x_0$ in dom(f) if and only if for each $\epsilon \gt 0$ there exists δ>0 such that x ∈dom(f) and $$
|x−x_0|<δ \implies |f(x)−f(x_0)| < \epsilon
$$I will, I think, benefit more if I shall be given a model proof.

Most introductory textbooks would have several of them. This is a PDF with a couple of examples. (I don't know if the link will download a PDF or not)