Can a continous interval have a finite number of points?

[Guest abcd123]

For example, is the interval [0.9999999...., 1] continous? Does it even make sense to ask that question?

[fourier jr]

I think only a function can be continuous. Maybe what you mean is 'is a point connected' (which is just what it sounds like). In that case, a point like [0.9999999..., 1] is connected.

[Dave]

Does it even make sense to ask that question?

 

Not as far as I'm aware. The issue of continuity seems to apply only to functions.

[Guest abcd123]

I asked that because the epsilon-delta definition of a limit does not appear to apply for the function defined for three consecutive values of x. For the function

 

f(x) =

.9999.... if x = .999999....

1 if x = 1

1.000...1 if x = 1.00000...1

 

the limit as x approaches 1 appears to intuitively be 1 (at least according to my intuition). However, the epsilon-delta definition of a limit (as presented by James Stewart's Calculus) demands that f be defined on an open interval containing 'a' (1 in this case) in order for the epsilon-delta definition to be applicable. What, then, is the limit as x approaches 1 in the function above?

 

I asked the question about continuity because if the interval (as defined in a function) wasn't continuous, then taking the limit would not make sense intuitively.

 

Any thoughts?

[MandrakeRoot]

Hi,

 

I dont understand your definition of your function f ?

If with 0.999999... you mean that this series of nine's continues indefinitely then it is the same point as 1, similarly if 1.000000000...1,means that you let the zeros continue indefinitely, then this is also 1. So you defined a function on a single point

{1}, and well if you see f as a function from {1} into R, where you let {1} inherit the metric of R restricted to {1} it is a continuous function, though not very interesting.

Remark that {1} is open in {1}. If you would consider the function f : [1 - eps,1+eps], with any eps > 0, defined by f(x) = x; then you could consider the restriction of f to (1 - 0.5eps,1+0.5eps) to have a(n) (open) neighbourhood of 1, and your function will equally be continuous.

 

Mandrake

[John]

Basically, given the information, the function is:

 

f(x) = x

 

on the interval [.999..., 1.000...1]. I know that's not exactly what abcd said, but we're dealing with points infinitely close to each other, so for all intents and purposes, that is the function.

 

lim x->n x

 

will always be n.

 

 

I asked the question about continuity because if the interval (as defined in a function) wasn't continuous, then taking the limit would not make sense intuitively.

 

If a function isn't continuous at a certain point, then the limit as the function approaches that point doesn't exist. You can work out the limit from either the left or the right, but since they won't be equal, by definition, the limit won't exist.

 

Those are my thoughts.

 

Edit: I forgot the answer the original question of the topic, heh. The answer is no. On a continuous interval, no matter which two points you choose, there will always be some point between them.

[Guest abcd123]

I may have incorrectly stated what I meant above. Also, forget what I asked about continuity.

 

Suppose a function, f, is defined on only 3 values of x that are infinitely close to each other. My question is what is the limit as x approaches the x-value of the middle point. It seems to me that it should be f(the middle point), but according to the epsilon-delta definition of a limit, taking the limit only makes sense on an open interval. f, being defined on only 3 points infinitely close to each other appears to me to be defined on a closed interval. So, what is the limit as x approaches the middle point? If the answer is that the limit doesn't exist, what is the reason behind a limit existing only on an open interval?

[Aeschylus]

I may have incorrectly stated what I meant above. Also' date=' forget what I asked about continuity.

 

Suppose a function, f, is defined on only 3 values of x that are infinitely close to each other. My question is what is the limit as x approaches the x-value of the middle point. It seems to me that it should be f(the middle point), but according to the epsilon-delta definition of a limit, taking the limit only makes sense on an open interval. f, being defined on only 3 points infinitely close to each other appears to me to be defined on a closed interval. So, what is the limit as x approaches the middle point? If the answer is that the limit doesn't exist, what is the reason behind a limit existing only on an open interval?[/quote']

Yes but there's no such thing as 'infinitely close', in the reals if the values are not the same as each other than there are an infinite number of reals between them.