Unclear About Proof

  • #1
http://i.imgur.com/uhqc5.jpg

I am trying to make sense of this proof in my textbook and find myself confused by a few things. I will try to present my question in the order that they come to me as I read the passage:

0) When defining the set A, "f is bounded above on [a,x]", x serves as a placeholder for whatever element x in [a,b] which, when evaluated by f, has the largest value? It is possible that this largest value can even be attained at x=b?

1) "A is bounded above (by b)"; I interpret this as meaning that b is just the greatest element of the domain; I'm guessing that this is relevant only because we need a nonempty set that has a greatest element, to be able to conclude that it has a least upper bound?

2) When the author indicates that the term "bounded above" refers to both the set A and to the function f, is this the case for all instances of the term, or is there one in particular that is relevant?

3) About f being bounded above (on the y-axis), the set {f(y): a <= y <= x} is represented this way to highlight that f(x) is the bound - whatever x may be?

4) When we suppose that b > alpha, this means that the purported LUB alpha is located in the continuous interval [a,b], we we can apply Theorem 1 (this is not included in what I posted, but it basically says, as you can infer, that if f is continuous, then there is a delta-interval where f is bounded above), by virtue of its continuity at this point.

5) At this point, I become less certain. We basically conclude from the above fact that there must be a number x1 on which f is bounded, so f is bounded on [a,x1]. This somehow contradicts that alpha is an upper bound for A. I understand why we can say there is a delta-interval about x1 where f is bounded, but why extend [a,x] to [a,x1]?

I'm not completely lost, it's just that I have difficulty expressing why this is so, and I think it's because the author's definition of the set A is confusing me. I will try to put it into words: If we choose our LUB (alpha) to be the greatest x-element, then if it is less than b (it must be continuous, since it is within a continuous interval) and therefore there is a delta-interval about it where it is bounded. So you can choose (by continuity) an element x1 within a positive delta distance from alpha, where the function is bounded. So you can extend the interval where the function is bounded; but this is a contradiction since we chose alpha as being the greatest element x element so you cant extend the interval further.

My confusion, I think, stems from the use of [a,x] in the definition; I believe the "x" is the largest value of the function itself, so when I follow the definition, I don't know why we can extend our interval to [a,x1] because of the fact that we chose an x1 greater than alpha; it is not alpha that is the greatest value of the function. I'll stop trying to rationalize here, since even as I write this I can see that I'm wrong.

6) I'm assuming the possibility that alpha > b can be disregarded, since we're only considering all x in [a,b]?

I'm sorry if this is lengthy to read. I would appreciate any kind of response that addresses any of my problems. It may suffice it anyone could explain the theorem concisely, perhaps in reference to the figure in my image, as I think I lack a geometric understanding of this. Thanks in advance.
 

Answers and Replies

  • #2
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0) Yes, in fact we will proof that b is always the largest value of the set A.

1) Yes. It means that every element in A is smaller (or equal) to b. Which is trivially true since [tex] A\subseteq [a,b] [/tex].

2) The author just wishes to indicate that there are two different notions of "bounded above".
One such notion is for sets and one such notion is for functions. A function is bounded above if there exists a c such that f(y)<c for every y (that is, f does not tend to infinity, and f does not have asymptotes). A set is bounded above if there exists a value greater then every element in the set.

3) This is probably the main source of your confusion... If we say that f is bounded above on [a,x], this just means that there exists a value C such that f(y)<C for every [a,x]. (You can illustrate this by drawing the horizontal line y=C. A function is bounded above by C is f always stays under the line). It does NOT mean that f(x) is the largest value.

For example, take the function f(x)=-x^2 on the interval [-1,1]. This function is bounded above on this interval. (You can draw the line y=1, f always stays below this line). However f(1) is not the maximal value.

A function which is not bounded above if it has an asymptote. So for example f(x)=1/x will not be bounded above on [-1,1]. Alternatively, there will not exist a C such that 1/x<C for every x in [-1,1]. (Note that f is not continuous on [-1,1], so f does not contradict the theorem)
So informally, you could say that a function is bounded above, if it does not have the behavior of 1/x.

4) Not really a question. But yes, you are right.

5) Instead of reading "f is bounded above", you could read "f does not have an asymptote" or "f stays below a certain horizontal line". Does it makes sense then?
Try to perform all the steps for the function f(x)=x^2 on [-1,1]. Where you presume that alpha=0. The proof then states clearly why alpha should not be 0.

6) Yes
 

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