Muddiest Point 10/24
One of the things we did in the class was proving Theorem 4.4.7, the Uniform Continuity on Compact Sets. Theorem 4.4.7 states that a function is continuous on a compact set K is uniformly continuous on K.
To me the prove in the class was hard to follow because the definition of compact set was not clear to me at that time. So I will review the concept of compact set and the prove again.
A set is compact if every sequence has a convergent subsequence converging in K. And another important theorem about compact set is Theorem 3.3.8 that states that the below statements are equivalent:
- K is compact.
- K is closed and bounded.
- Every open cover of K has a finite subcover.
The third statement is a little bit unfamiliar to us but the first two statements are very useful to us. And with that we can easily prove a continuous function f on a bounded subset of is uniformly continuous because a closed and bounded set is compact and then we can use Theorem 4.4.7 to prove that it is uniformly continuous.
The prove of Theorem 4.4.7 proceeds with proving by contradiction. Since we want to prove that is uniformly continuous, the contradiction to this is to prove that there exist two sequences and in K such that lim yet for some .
Here we will utilize the concept of K being a compact set such that the sequence has a
convergent subsequence $latex(x_{n_k})$ with x = lim also in K. Now we can use the Algebria Limit Theorem such that lim lim . So now we find a contradiction that lim contradicting with for some . Thus we can prove that f is uniformly continuous on K.
This prove utilizes different Theorem that we learned in the previous classes so it is helpful that we can review the old concepts that we learned.
Good post Zehao. I am happy that you are explicitly going back and reviewing material from previous courses. This activity is very important, and I hope that everyone takes the time to flip back through what we have proven periodically, to ensure that they are familiar with all of the important results.