Continuity

[Ethan3141]

Hey so basically I have to show that [MATH]f(x) = \begin{cases} x & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}/\mathbb{Q} \end{cases} [/MATH]Is only continuous at x=0.
There are a couple of things bothering me about this question:
Firstly, I prove the function isn't continuous at [MATH]x\neq0[/MATH] but to do this I split it into two cases: [MATH]x \in \mathbb{Q}/\mathbb{0}[/MATH] and [MATH]x \in \mathbb{R}/\mathbb{Q}[/MATH] (The method I use in both is the same so I will just show the working for one case)
Case 1: [MATH]x \in \mathbb{Q}/\mathbb{0}[/MATH]: Let [MATH](x_n)_n[/MATH] be a sequence of irrational numbers converging to [MATH]a[/MATH]Then [MATH]f(x)=x[/MATH] but [MATH]\lim_{n\to\infty}{f(x_n)}=0[/MATH] Thus [MATH]f(x)[/MATH] is not continuous when [MATH]x \in \mathbb{Q}/\mathbb{0}[/MATH]
Question 1: Is this method correct?
Question/Clarification 2: How does the limit of the sequence not being equal to value of the function at that point mean it is not continuous? The lecturer kept referring to this fact in our seminar but I don't think he has introduced it or if he has then it made no sense to me.
Question 3: Is there a shorter way of writing "Let [MATH](x_n)_n[/MATH] be a sequence of irrational numbers converging to [MATH]a[/MATH]"?
Maybe [MATH](x_n)_n\subseteq(\mathbb{Q}/\mathbb{0})\longrightarrow a[/MATH]?
Any help would be greatly appreciated.

Thanks, Ethan

 

[pka]

Hey so basically I have to show that [MATH]f(x) = \begin{cases} x & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}\setminus\mathbb{Q} \end{cases} [/MATH]Is only continuous at x=0.
There are a couple of things bothering me about this question:
Firstly, I prove the function isn't continuous at [MATH]x\neq0[/MATH] but to do this I split it into two cases: [MATH]x \in \mathbb{Q}\setminus\mathbb{0}[/MATH] and [MATH]x \in \mathbb{R}/\mathbb{Q}[/MATH] (The method I use in both is the same so I will just show the working for one case)
Case 1: [MATH]x \in \mathbb{Q}/\mathbb{0}[/MATH]: Let [MATH](x_n)_n[/MATH] be a sequence of irrational numbers converging to MATH]a[/MATH]
Then [MATH]f(x)=x[/MATH] but [MATH]\lim_{n\to\infty}{f(x_n)}=0[/MATH] Thus [MATH]f(x)[/MATH] is not continuous when [MATH]x \in \mathbb{Q}\setminus\mathbb{0}[/MATH]Question 1: Is this method correct?
Question/Clarification 2: How does the limit of the sequence not being equal to value of the function at that point mean it is not continuous? The lecturer kept referring to this fact in our seminar but I don't think he has introduced it or if he has then it made no sense to me.
Question 3: Is there a shorter way of writing "Let [MATH](x_n)_n[/MATH] be a sequence of irrational numbers converging to [MATH]a[/MATH]"?
Maybe [MATH](x_n)_n\subseteq(\mathbb{Q}/\mathbb{0})\longrightarrow a[/MATH]?
Any help would be greatly appreciated.

First note that the use of \setminus is proper here [ tex]\mathbb{R}\setminus\mathbb{Q} [/tex] gives \(\displaystyle \mathbb{R}\setminus\mathbb{Q}\)
Suppose that \(\displaystyle \mathop {\lim }\limits_{x \to {x_0}} f(x) = L\) then if \(\displaystyle \left( {{x_n} \to {x_0}} \right) {\bf\text{ implies }} \left( {f({x_n}) \to L} \right)\)
But I don't think that is what you want, Suppose that \(\displaystyle a\in\mathbb{R}\setminus\mathbb{Q}\) that is \(\displaystyle a\) is an irrational number.
Now we know that \(\displaystyle a\ne 0\) WHY? So that \(\displaystyle |a|>0\). We also know \(\displaystyle (q_n)\to a\) there is a sequence of rational numbers that converge to \(\displaystyle a\) Now \(\displaystyle (\forall n)[f(q_n)=0]\) Can you explain that? Can \(\displaystyle f(q_n)\to f(a)=a~?\)
So please post any questions you still have?

 

[pka]

EDIT
Now we know that \(\displaystyle a\ne 0\) WHY? So that \(\displaystyle |a|>0\). We also know \(\displaystyle (q_n)\to a\) there is a sequence of irrational numbers that converge to \(\displaystyle a\) Now \(\displaystyle (\forall n)[f(q_n)=0]\) Can you explain that? Can \(\displaystyle f(q_n)\to f(a)=a\ne 0~?\)
So please post any questions you still have?

 

[Ethan3141]

First note that the use of \setminus is proper here [ tex]\mathbb{R}\setminus\mathbb{Q} [/tex] gives \(\displaystyle \mathbb{R}\setminus\mathbb{Q}\)
Suppose that \(\displaystyle \mathop {\lim }\limits_{x \to {x_0}} f(x) = L\) then if \(\displaystyle \left( {{x_n} \to {x_0}} \right) {\bf\text{ implies }} \left( {f({x_n}) \to L} \right)\)
But I don't think that is what you want, Suppose that \(\displaystyle a\in\mathbb{R}\setminus\mathbb{Q}\) that is \(\displaystyle a\) is an irrational number.
Now we know that \(\displaystyle a\ne 0\) WHY? So that \(\displaystyle |a|>0\). We also know \(\displaystyle (q_n)\to a\) there is a sequence of rational numbers that converge to \(\displaystyle a\) Now \(\displaystyle (\forall n)[f(q_n)=0]\) Can you explain that? Can \(\displaystyle f(q_n)\to f(a)=a~?\)
So please post any questions you still have?

Which question is this an answer to? I assume it is question 1 or question 2.

Question 1: Is this method correct?
Question/Clarification 2: How does the limit of the sequence not being equal to value of the function at that point mean it is not continuous? The lecturer kept referring to this fact in our seminar but I don't think he has introduced it or if he has then it made no sense to me.
Question 3: Is there a shorter way of writing "Let [MATH](x_n)_n[/MATH] be a sequence of irrational numbers converging to [MATH]a[/MATH]"?
Maybe [MATH](x_n)_n\subseteq(\mathbb{Q}/\mathbb{0})\longrightarrow a[/MATH]?

Anyway regarding this:

Can you explain that? Can \(\displaystyle f(q_n)\to f(a)=a\neq0~?\)

Considering [MATH]q_n[/MATH] is a sequence of irrational numbers then by definition of the function, like you stated: \(\displaystyle (\forall n)[f(q_n)=0] \) but \(\displaystyle f(q_n)\to a\) But isn't this just the irrational case of my proof? Anyway one question I have about this is does this imply that \(\displaystyle \mathop {\lim }\limits_{x \to {x_0}} f(x)\) does not exist or just that it isn't equal to [MATH]f(x_0)[/MATH]

I split it into two cases: [MATH]x \in \mathbb{Q}/\mathbb{0}[/MATH] and [MATH]x \in \mathbb{R}/\mathbb{Q}[/MATH] (The method I use in both is the same so I will just show the working for one case)

Sorry if I misunderstood but I am just confused. Clarification on what question you were answering would be really helpful to me. Regardless thank you so much for the help.

Thanks, Ethan

 

[pka]

Which question is this an answer to? I assume it is question 1 or question 2.
Considering [MATH]q_n[/MATH] is a sequence of irrational numbers then by definition of the function, like you stated: \(\displaystyle (\forall n)[f(q_n)=0] \) but \(\displaystyle f(q_n)\to a\) But isn't this just the irrational case of my proof? Anyway one question I have about this is does this imply that \(\displaystyle \mathop {\lim }\limits_{x \to {x_0}} f(x)\) does not exist or just that it isn't equal to [MATH]f(x_0)[/MATH]Sorry if I misunderstood but I am just confused. Clarification on what question you were answering would be really helpful to me. Regardless thank you so much for the help.

1) Are you clear that the function is continuous at \(\displaystyle x=0~?\)
2) I was then looking at the case \(\displaystyle x\ne 0\). That is what the proof I posted is about.

 

[Ethan3141]

1) Are you clear that the function is continuous at \(\displaystyle x=0~?\)
2) I was then looking at the case \(\displaystyle x\ne 0\). That is what the proof I posted is about.

Thanks for the clarification, I am clear that the function is continuous at [MATH]x=0[/MATH]. My main issue was the reason the method works but I have since went back to using the Epsilon-Delta method as I understand it more.

 

[pka]

Thanks for the clarification, I am clear that the function is continuous at [MATH]x=0[/MATH]. My main issue was the reason the method works but I have since went back to using the Epsilon-Delta method as I understand it more.

This proof is an example where notions of neighborhoods from a metric space are really useful.
But if you have more questions please post as clear a description as possible.

 

[Ethan3141]

This is my full solution. 6.24 references a proposition from our lecture in which we were shown that [MATH]\mathbb{Q}\subseteq\mathbb{R}[/MATH] and [MATH]\mathbb{R}\setminus\mathbb{Q}\subseteq\mathbb{R}[/MATH] are dense. I am aware there are some statements missing, eg: Thus ... is discontinuous at ... and thus ... is only continuous at ... (This was just a rough draft)

Questions:
1) I am not sure that case 1 is correct, I tried all I could think of (reverse triangle inequality) to get the desired form to prove something isn't continuous. Any pointers on where I have gone would be greatly appreciated

Thanks, Ethan