Rigorous definition of continuity on an open vs closed interval

[kahwawashay1]

Let I be an open interval and f : I → ℝ is a function. How do you define "f is continuous on I" ?

would the following be sufficient? :

f is continuous on the open interval I=(a,b) if \stackrel{lim}{x\rightarrow}c \frac{f(x)-f(c)}{x-c} exists \forall c\in (a, b)

is this correct?

Also, what about the case of a closed interval I? In that case, can you just add to the above statement that:

\stackrel{lim}{x\rightarrow}a^{+} f(x) = f(a)
and
\stackrel{lim}{x\rightarrow}b^{-} f(x) = f(b)

?

 

[dextercioby]

The definition you gave is for differentiability on an open interval. The definition for continuity is with delta and epsilon. The [ itex ] for lim looks like \displaystyle{\lim_{x\rightarrow a}}.

 

[kahwawashay1]

The definition you gave is for differentiability on an open interval. The definition for continuity is with delta and epsilon. The [ itex ] for lim looks like \displaystyle{\lim_{x\rightarrow a}}.

But differentiability implies continuity?
Anyway, then would this be right:

f is continuous on the open interval I=(a,b) if |\frac{f(x)-f(c)}{x-c} - f^'(c)|< ε when |x-c|<δ \forall c\in (a, b) and what about my use of the right and left hand limits for the case of a closed interval [a,b]? would that be correct?

 

[dextercioby]

[...]
f is continuous on the open interval I=(a,b) if |\frac{f(x)-f(c)}{x-c} - f^'(c)|< ε when |x-c|<δ \forall c\in (a, b) [...]

I was referring to the http://en.wikipedia.org/wiki/Continuous_function Weierstrass epsilon-delta definition for continuity.

 

[kahwawashay1]

No, f is differentiable on the open interval I=(a,b) if |\frac{f(x)-f(c)}{x-c} - f^'(c)|< ε when |x-c|<δ \forall c\in (a, b) [...]

ohhhh nvm. it can be continuous on (a,b) but not differentiable, like the abs value of x.
Ok then it would just be:
f is continuous on (a,b) if \displaystyle{\lim_{x\rightarrow c}} f(x) = f(c) \forall c\in (a, b)

right?

 

[dextercioby]

Yes, but for that limit to make sense, you have to use the topology, that is compare values of f(x_1) and f(x_2) when x_1 and x_2 get arbitrarily close to each other. That's what the epsilon-delta definition does.

 

[kahwawashay1]

Yes, but for that limit to make sense, you have to use the topology, that is compare values of f(x_1) and f(x_2) when x_1 and x_2 get arbitrarily close to each other. That's what the epsilon-delta definition does.

we didnt learn about topology yet so idk what is f(x_1) and the like...