Isolated points and continuity

[dancergirlie]

Homework Statement

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

Homework Equations

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!

 

[tiny-tim]

Hi dancergirlie!

(have a delta: δ and an epsilon: ε )

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.

How can that statement not be true?

 

[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.

 

[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 [itex]\delta[/itex] such that [itex](c-\delta,c+\delta)[/itex] contains no other points of A. Hint: that is your [itex]\delta[/itex] to show that f is continuous at c. So now let x be in A and [itex]\varepsilon>0[/itex]. Then...

 

[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?

 

[n!kofeyn]

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?

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 [itex]|x-c|<\delta[/itex] imply if x is in A?

 

[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...

 

[n!kofeyn]

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...

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

 

[tiny-tim]

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.

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? ​

 

[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..

 

[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?

 

[n!kofeyn]

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

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

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

 

[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!

 

[n!kofeyn]

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

 

[tiny-tim]

I understand that I can find a delta for any specific epsilon …

ok, then find it for ε = 2009

… but how am I supposed to phrase that in my proof for an arbitrary epsilon?

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

 

[HallsofIvy]

Take [itex]\delta[/itex] 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 [itex]0< |x-c|< \delta[/itex] so the hypothesis, "if [itex]0< |x-c< \delta[/itex]" is always false. If the hypothesis of a statement is false then the statement is _____.