Thread: f continuous on (a, b) extended to continues on [a, b] imply f is uniformly continues

[FONT=&quot] If fa, b) -->R is continuous and it can be extended to be continuous on [a, b], prove that f is uniformly continuous.

I think there is a hole in my logic for solving this problem with the proof idea I have. My first though is:

Since f is continues, we can create a ball of radius (epsilon) around every f(x) belonging to
f([a, b]) with x belonging to (a, b). f being continuous means that each ball of f(x) has a corresponding (Delta) which creates a ball around x of radius (delta). Choose the smallest delta. Using this (delta), one can show this function is uniformly continuous.

The problem I have with this sketch of a proof is that I don't use the fact [a, b] is closed. I could use [a, b] being compact to give myself a limited number of balls, but seeing as the set of all delta's exist in the real number system which has the least upper bound property, I'm not sure it needed. Secondly I don't use the fact that the domain is extended from (a, b) to [a, b]. I had a second idea dealing with right handed and left handed limits, but it once again wouldn't use the domain being extended. Any help would be greatly appreciated!

I am not sure what to say here.
Every continuous function on a closed and bounded interval of real numbers is uniformly continuous. The standard proof uses the fact that such sets are compact. From any open cover there is a finite sub-cover.
So some of your ideas about a proof are correct.