You have some confused notation here.
You have in two different ways.
What you want is to note that is an upper bound for .
Now consider so that .
Then show that
I have attempted this proof and would appreciate any feedback. Thanks.
Let be some interval. If is increasing and , prove that is an upper bound for and .
Proof: Let be an increasing function. Then, by definition, for all . Then, let for and all . Thus, since for all . Hence, is an upper bound for .
Then, since for all , then must be the least upper bound for . Hence, .
Should be f(a), not a.Hence, is an upper bound for .
The second claim is not really proved.
Even the claim that f(a) (rather than a) is the least upper bound for is incorrect. ConsiderThen, since for all , then must be the least upper bound for .
Then , but .
Even if , how does it help prove the following?After all, only if f is left-continuous at a. In my opinion, there is at least not enough details here that are expected from a proof of such a simple fact.Hence, .
You can't expect that x < a for all x in I. Since a is already fixed in the problem statement, this phrase should say, "Let x < a for some x in I".Then, let for and all .
Again, it is not the case that f(x) < f(a) holds for all a, x in I.Thus, since for all .
Alright, so from what I gather, most of the first part was ok besides the fact that is the upper bound and not .
As for the second part...
Let and . Then,
since is an upper bound for and f is increasing then there is no such that . Thus, . If we let this limit be called then we have that where . Hence, we have that .
I'm not sure this is very strong but its my second attempt.
Sorry, too many flaws. The second part of the problem -- proving -- has nothing to do with f(a). In the example I gave above, f(a) is different from .
Are you assuming ? In this case, since this assumption is not in the problem statement, you later need to consider the possibility that . However, I thought that you have proved this in part 1. Therefore, you should say, "Let . We have already shown that ." (To repeat, the second sentence is irrelevant.)Let and .
What happens to the right of is irrelevant to the problem.Then, since is an upper bound for and f is increasing then there is no such that .
I don't see how this follows from the previous discussion.Thus, .
This is redefining A.If we let this limit be called
It looks like this conclusion is based on confusing two meanings of A.then we have that where . Hence, we have that .
Let us define , as you have done. Here is how I would prove that . We show that for every there exists a such that when and we have . Note that the second inequality, i.e., , holds by the definition of .
Let be given. Also by the definition of , is not an upper bound of ; therefore, there exists an such that . But then, since f is increasing, for every such that we have . Therefore, we can take .
A quasi-continuous function is one for which the right and left hand limits exist at each point. It is a standard theorem in analysis that any monotonic function is quasi-continuous. That is the essence of this problem. You are asked to show that the limit on the left exist at each point.