Advanced Calculus/Real Analysis Help

Hello Everyone! I have 2 questions that I am extremely stuck on and I was wondering if anyone knew how to help! Anything would be great. Here they are:

(1) Use Bolzano Weierstrass to prove the intermediate value theorem. "If f is a continuous function on a,b and f(a) & f(b) have opposite signs, then there is a point c on (a,b) such that f(c)=0.

(2) Prove that every convergent sequence has an increasing or decreasing subsequence, or maybe both. Do so using the limi as n goes to infinity of x(sub n) is equal to l.

Thank you!

Re: Advanced Calculus/Real Analysis Help

Hi,

Problem 1.

Let f be a continuous function on the interval I, [x,y] with [x,y] subset of I, and t $\displaystyle \in$ R with f(x) < t < f(y), then there exists s $\displaystyle \in$ (x,y) s.t f(s) = t.

W.L.O.G suppose that f(x) < t < f(y).

Let U = {u $\displaystyle \in$ (x,y) | f(u) < t}, x $\displaystyle \in$ U, so U is non-empty.

Clearly, the set U is bounded above by y.

Let s = sup U, and n $\displaystyle \in$ N. Since s is the sup of U, s - 1/n is not an upper bound for U,

By bolzano-weierstrass, there exists a x_n $\displaystyle \in$ U $\displaystyle \cap$ such that s - 1/n < x_n < s. For all n $\displaystyle \in$ N, x_n $\displaystyle \in$ U implies that f(x_n) < t. (this is how U is defined)

You can see that x_n converges to the sup U = s. By continuity, f(x_n) converges to f(s) $\displaystyle \leq$ f(t).

Finish the proof, using a similar argument, you want to show that f(s) $\displaystyle \geq$ f(t).

Hint: Use Bolzano-Weirstrass, the fact that s = sup U, and the fact that there is a convergent subsequence such that y_n is not in U.

Problem 2

We first need to define a dominant term. The nth term of a sequence is dominant if for all m > n, a_m < a_n. In other words the dominant term is greater than all of the elements that come after it. (note that this term may be defined a little different in your book or an inferior term may be discussed). Anyway, the proof would not be any different.

Now, we continue our proof by inspecting two cases.

1. First case: A sequence contains infinitely many dominant terms.

-Find the first dominant term, then the second, then the third. Do you see a pattern? You can obtain a decreasing subsequence by listing all of the dominant terms in order.

2 Second case: A sequence contains finitely many dominant terms.

- List all of the dominant terms. Then, select the element a_k that comes right after the last dominant term, and construct your increasing subsequence with those elements. Do you see why?

Let me know if you run into problems.

Re: Advanced Calculus/Real Analysis Help

Thank you so much for your response, I really appreciate you taking your time out and helping me. For number 1 I'm not really sure how to use sup as my class has not gotten to it yet. For number 2 what exactly do you mean by dominant term?

Thank you!

Re: Advanced Calculus/Real Analysis Help

The supremum is the least upper bound of a set.

What do we mean by a least upper bound?

Def'n:

Upper Bound. Let S be a subset of R. If there exists a M $\displaystyle \in$ R such that for all x $\displaystyle \in$ S, one has x$\displaystyle \leq$ M, then M is an upper bound for S.

In other words, an upper bound for a set is a number greater than any element of the set.

Take the set A ⊆ R, A = {1,2,3}. Well, 4, by definition, is an upper bound of this set, right? 4 is greater than any element in the set A. Similarly, 3.000001 and 7/2 are upper bounds. 3 is also an upper bound, in particular the least upper bound, since 3 is greater than or equal to all of the elements in that set and less than any other upper bound.

Def'n:

Least upper bound, supremum: Let S be a subset of R. Suppose S is bounded above (has an upper

bound). If there exists s ∈ R such that s is an upper bound for S and for all upper bounds

M of S, one has s ≤ M, then we call s the least upper bound or supremum of S, denoted

supS.

In other words, the least upper bound is the smallest of the upper bounds.

Take the set T = { x | x $\displaystyle \in$ R, x $\displaystyle \leq$ 1}. What are some upper bounds for this set? What is the least upper bound of this set? Is the least upper bound of this set T in the set or not?

What is the supremum of the empty set? What can you say about this?

Note: Don't forget to look for the definition of infimum. An infimum is the greatest lower bound of a set (or a sequence), depending on the context. It is defined similarly.

Also, the existence of supremums and infimums relies on the Completeness Axiom. This is crucial in Real Analysis.

"Let S ⊆ R be nonempty and bounded above. Then, supS ∈ R." A lot of your proofs will revolve around the idea of having a supremum (or infimum). Depending on your analysis course, you may be asked to prove that the set of the reals is complete. In any case, like I said, this idea is fundamental in Real Analysis.

Moving on to dominant terms, let a_n = 1 / n

a1 = 1, a2, = 1/2... This sequence has infinitely many dominant terms. Why? 1 > 1/2, 1/2 > than all of the terms after it. Any term in the sequence is greater than all of the terms following it. Hence, every term of this sequence is a dominant term.

Let's construct a sequence say: 1,4,2,4, 1, 2, 1, 4, 5, 7, 9, 0, 0, 0, 0, 0... (all zeroes from this point on).

The dominant term of this sequence is 9. Why? Look at the definition. Why is 1 not a dominant term? Are there finitely many or infinitely many dominant terms?

Let me know if you have further questions!