Extending a uniformly cont function on an open interval to a closed interval?

[moxy]

Homework Statement

Show that the function [itex]f: J → ℝ[/itex] is bounded if [itex]f[/itex] is uniformly continuous on the bounded interval J.

Homework Equations

J is a bounded interval, so say J = (a,b)

[itex]f[/itex] is uniformly continuous on J, so
[itex]\forall \epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that for [itex]s,t \in J = (a,b)[/itex]
[itex]|f(s) - f(t)| < \epsilon[/itex] whenever [itex]|s - t| < \delta[/itex]

[itex]f: J → ℝ[/itex] is bounded if there exists a real number M such that |f(x)| ≤ M for all x in J.

The Attempt at a Solution

I think that if I can extend [itex]f[/itex] to the endpoints of J, then I can use the Extreme Value Theorem to say that [itex]f[/itex] attains a min and max value, i.e. is bounded. So I need to define [itex]f(a)[/itex] and [itex]f(b)[/itex].

[itex]f(a) = \lim_{n→∞}{f(a + \frac{1}{n})}[/itex]
[itex]f(b) = \lim_{n→∞}{f(b - \frac{1}{n})}[/itex]

Where clearly [itex]f[/itex] is defined on [itex][a + \frac{1}{n}, b - \frac{1}{n}][/itex] for all [itex]n \in ℕ[/itex]


Am I on the right track? I don't feel like I've used the fact that [itex]f[/itex] is uniformly continuous on J. Is it because [itex]f[/itex] is uniformly continuous that I'm able to define [itex]f(a)[/itex] and [itex]f(b)[/itex]?

 

[deluks917]

Try to do this more simply. The idea is to pick x1,...,xn such that every x in J is "close" to one of these xi. This plus uniform continuity will give you what you want.

 

[moxy]

But I only know that J is bounded, not that it's closed and bounded. So can't I only say that J = (a,b) and not necessarily J = [a,b]? So I don't immediately know if f is defined at a or b, much less if it's continuous at a and b.

 

[moxy]

Okay, nevermind, you changed your post. So... I should define a sequence of, say, rational numbers {x_n}, where all x_i are in J, such that [itex]lim_{n->∞}{\{x_n\}} = a[/itex] ?

And similarly, {y_n}, where all y_i are in J, such that [itex]lim_{n->∞}{\{y_n\}} = b[/itex] ?

I think that I'm able to do this, but I don't see how it gets me any closer to showing that f is bounded.

EDIT:

I guess I can say that

[itex]lim_{n->∞}{f(x_n)} = f(a)[/itex]
[itex]lim_{n->∞}{f(y_n)} = f(b)[/itex]

So f is defined on [a,b]. Do I still have to show it's continuous at a and b to use the extreme value theorem? I don't see how this approach is simpler or much different than the one in my original post.