# Uniform Continuity Proof 2

• Oct 15th 2007, 02:59 PM
tbyou87
Uniform Continuity Proof 2
Let f be a function on (a,b). Suppose that f is uniformly continuous on (a,c] and [c,b) for some c such that a<c<b. Then, Prove that f is uniformly continuous on (a,b) by using the epsilon - delta definition.

My first problem is that I don't get if i need to have epsilon1 and epsilon2 or delta1 or delta2 for each uniformly continuous function or if i can call them both by the same variable. I'm assuming you need to combine both definitions somehow to get it over the bigger interval.

Thanks
• Oct 16th 2007, 03:08 PM
ThePerfectHacker
Quote:

Originally Posted by tbyou87
Let f be a function on (a,b). Suppose that f is uniformly continuous on (a,c] and [c,b) for some c such that a<c<b. Then, Prove that f is uniformly continuous on (a,b) by using the epsilon - delta definition.

My first problem is that I don't get if i need to have epsilon1 and epsilon2 or delta1 or delta2 for each uniformly continuous function or if i can call them both by the same variable. I'm assuming you need to combine both definitions somehow to get it over the bigger interval.

Thanks

Theorem: A function is uniformly continous on \$\displaystyle (a,b)\$ if and only if it can be enlarged to a continous function on \$\displaystyle [a,b]\$.

Now let \$\displaystyle f_1\$ be the uniformly continous function on \$\displaystyle (a,c]\$ and \$\displaystyle f_2\$ be the uniformly continous function on \$\displaystyle [c,b)\$. We want to show that \$\displaystyle f\$ which is the peices of \$\displaystyle f_1,f_2\$ combined together is uniformly continous on \$\displaystyle (a,b)\$. Now by theorem we can enlarge \$\displaystyle f_1\$ to \$\displaystyle g_1\$ to the closed interval \$\displaystyle [a,c]\$ and we can enlarge \$\displaystyle f_2\$ to a continous function \$\displaystyle g_2\$ on \$\displaystyle [c,b]\$. Then define \$\displaystyle g\$ as the function obtained from combining \$\displaystyle g_1,g_2\$ on the full interval \$\displaystyle [a,b]\$. Using the theorem the continous functions on closed finite intervals are uniformly continous it means \$\displaystyle g\$ is uniformly continous on \$\displaystyle [a,b]\$ which means it is certainly uniformly continous on a smaller interval \$\displaystyle (a,b)\$ which is \$\displaystyle f\$.
• Oct 16th 2007, 04:06 PM
Plato
While that proof works, I wonder how many basic analysis course prove an enlargement theorem.

The classic proof of this theorem simply requires three cases.
The only one of which that cases any concern is \$\displaystyle x < c < y\$.
• Oct 16th 2007, 04:07 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
While that proof works, I wonder how many basic analysis course prove an enlargement theorem.

My analysis class omitted that theorem. But I learned it from the book anyway. It is easy, I do not know why omit it.
• Oct 16th 2007, 04:21 PM
Plato
Quote:

Originally Posted by ThePerfectHacker
My analysis class omitted that theorem. But I learned it from the book anyway. It is easy, I do not know why omit it.

I think that can explain why some techniques are omitted.
Often, what appears to be an overly tedious proof actually teaches an important proof technique.
I really think that this is such an example.

Sometimes known theorems stand in the way of learning mathematics.
• Oct 16th 2007, 04:28 PM
ThePerfectHacker
Quote:

Originally Posted by Plato
Sometimes known theorems stand in the way of learning mathematics.

But you have to agree that is not always the case. Some theorems are so disatrous in proof that it is okay to omit them, i.e. many fundamental theorems are like that. Thus, no harm is done towards the student if such a proof is omitted. But in general most theorems are straightforward and can present an important technique which the student might learn from.

My approach to analysis have been like I described about. I try to first learn how to do computational analytical stuff first. Then once I mastered that I go into the theory.

But that is only in analysis. In algebraic theories I do not feel that this is the best approach.