# Isolated points and continuity

1. Nov 1, 2009

### dancergirlie

1. The problem statement, all variables and given/known data
Let f : A --> R be a function, and let c in A be an isolated point of A. Prove that f
is continuous at c

2. Relevant equations

3. The attempt at a solution

I'm kind of confused by this problem.... if c is an isolated point, then the limit doesn't exist. So I can't really use the fact that a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon.

Any hints would be great!

2. Nov 1, 2009

### tiny-tim

Hi dancergirlie!

(have a delta: δ and an epsilon: ε )
How can that statement not be true?

3. Nov 1, 2009

### dancergirlie

Well it wouldn't be true for x=c. Because |x-c| would be equal to zero, and that contradicts the statement that delta is greater than zero.

4. Nov 1, 2009

### n!kofeyn

I am assuming that A is a subset of the real numbers. Since c is an isolated point of A, there exists a $\delta$ such that $(c-\delta,c+\delta)$ contains no other points of A. Hint: that is your $\delta$ to show that f is continuous at c. So now let x be in A and $\varepsilon>0$. Then...

5. Nov 1, 2009

### dancergirlie

I am not sure how you are choosing delta.... are you saying that you choose delta to be equal to the delta satisfying (c-delta, c+delta) containing no other points of A but c?

6. Nov 1, 2009

### n!kofeyn

Yes. That delta exists by the fact that c is an isolated point. Now just go through the epsilon-delta steps of proving f is continuous at c. Using that delta, what does $|x-c|<\delta$ imply if x is in A?

7. Nov 1, 2009

### dancergirlie

Would it just mean that x-c is in a as well?
Because I'm trying to show that
|f(x)-L|< epsilon, but there is no limit so that is why I'm getting confused....

8. Nov 1, 2009

### n!kofeyn

Well your L is f(c). So you are trying to show that for x in A and any $\varepsilon>0$, there is a $\delta>0$ such that $|x-c|<\delta$ implies that $|f(x)-f(c)|<\varepsilon$. But if x is in A and $|x-c|<\delta$, where the delta is the one described above, then x can only be one point! And that point is ...?

9. Nov 1, 2009

### tiny-tim

Sorry, but that doesn't make sense.

How does any of that answer the question, can the statement "a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon" be untrue?

Choose any ε … for example choose ε = 2009 …

can you find a δ that works for that ε?

10. Nov 1, 2009

### dancergirlie

Wouldn't that mean that x has to be c??? If it was then that contradicts the fact that it is continuous since that would mean that epsilon would=0 and delta=0, so the function wouldn't be continuous.... but I'm trying to prove that it is continuous. I don't know if I m completely off here... but that is what I'm getting from what you're saying..

11. Nov 1, 2009

### dancergirlie

[How does any of that answer the question, can the statement "a function is continuous at c if for all epsilon>0 there exists a delta>0 such that whenever |x-c|<delta, it follows that |f(x)-f(c)|<epsilon" be untrue?

Choose any ε … for example choose ε = 2009 …

I understand that I can find a delta for any specific epsilon, but how am I supposed to phrase that in my proof for an arbitrary epsilon?

12. Nov 1, 2009

### n!kofeyn

Yes, it would mean that x=c, but that does in no way imply that $\varepsilon=0$. Look. I'll just rewrite everything so hopefully it makes sense now.

Let c be an isolated point of A. Then there exists a $\delta>0$ such that the interval $(c-\delta,c+\delta)$ contains no other points of A besides c.

Let x be in A and $\varepsilon>0$. Then $|x-c|<\delta$ implies that x is in the interval $(c-\delta,c+\delta)$. This means x=c because there are no others points of A in that interval. This implies that $|f(x)-f(c)|=|f(c)-f(c)|=0<\varepsilon$ for absolutely any $\varepsilon>0$ you choose. Therefore, f is continuous at c.

13. Nov 1, 2009

### dancergirlie

Oh it makes sense, I was thinking that implied that epsilon equals zero, when it was just that |f(x)-f(c)|=0 which would be less than any epsilon greater than zero.... thank you so much for your help, i really appreciate it!!

14. Nov 1, 2009

### n!kofeyn

Good! No problem for the help. This problem is subtle, so I'm glad it makes sense now.

15. Nov 1, 2009

### tiny-tim

ok, then find it for ε = 2009
well, for example, if you can use the same δ for every ε, that would do it, wouldn't it?

16. Nov 2, 2009

### HallsofIvy

Staff Emeritus
Take $\delta$ to be less than the distance from closest point in the domain of f other than c. There are no points in the domain of f such that $0< |x-c|< \delta$ so the hypothesis, "if $0< |x-c< \delta$" is always false. If the hypothesis of a statement is false then the statement is _____.