Sequence, continuity, connectedness

kingwinner
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1) Prove that
lim x_k exsts and find its value if {x_k} is defined by
k->inf
x_1 = 1 and x_(k+1) = (1/2) x_k + 1 / (sqrt k)
[My attempt: Assume the limit exists and equal to L
then L= (1/2) L + 0
=> (1/2) L = 0
=> L=0
Now I have to prove that the limit indeed exists, I want to use the monotone sequence theorem (monotone & bounded => convergence), but as I evaluate a few terms, I found that x1<x2, but x2>x3, which makes it not montone...what should I do?]

Definition: Let S be a subset of R^n. S is called open if it contains none of its boundary points. A point x E R^n is a boundary point of S if every ball centred at x contains both points in S and points in S^c (S complement)

Definition: A subset of R^n is called compact if it's both closed and bounded.

Definition: f is continuous at a iff
lim f(x) = f(a)
x->a

2) Suppose f: R^m->R^n is a map such that for any compact set K C R^n, the preimage set f^(<-)(K) = {x E R^m | f(x) E K} is compact. Is f necessary continuous? Justify.


3) Suppose f: R^n -> R^k has the following property: For any open set U C R^k, {x| f(x) E U} is an open set in R^n. Show that f is continuous on R^n.


For #2 and #3, I have absolutely no clue...feeling desperte...


Any help will be greatly appreciated!:smile:
 
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You must show that x is a monotone decreasing sequence. Since x_(k+1) = (1/2) x_k + 1 / (sqrt k), x_(k+1) would be < x_k for sufficiently small values of 1 / (sqrt k). You need to solve for that k. (You can discard the first few terms of the sequence if they are not monotone decreasing.)

What is your definition of an open set? A compact set?
 
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Q1. Perhaps x_2 > x_3 > x_4 > ...?

Q2. Do you think this is true or false? A continuous function necessarily pulls back closed sets to closed sets. By the Heine-Borel theorem, the compact subsets of R^n (and R^m) are precisely those that are closed and bounded. So f pulls back at least these subsets to closed subsets. But there are more closed sets out there (specifically the ubounded ones). Maybe you can use this to construct a counterexample?

Q3. What definition of continuity are you working with?
 
So f pulls back at least these subsets to closed subsets. But there are more closed sets out there (specifically the ubounded ones).
There is a direct, specific relationship between compact-to-compact and continuity.
 
morphism said:
Q1. Perhaps x_2 > x_3 > x_4 > ...?

Q2. Do you think this is true or false? A continuous function necessarily pulls back closed sets to closed sets. By the Heine-Borel theorem, the compact subsets of R^n (and R^m) are precisely those that are closed and bounded. So f pulls back at least these subsets to closed subsets. But there are more closed sets out there (specifically the ubounded ones). Maybe you can use this to construct a counterexample?

Q3. What definition of continuity are you working with?

1) Yes, I've found out that the pattern is x2 > x3 > x4 > x5 > ...

Now is it possible to show by math induction that x_k > x_(k+1) > 0 (hence bounded and montone) ?

2) My course didn't cover the Heine-Borel theorem, so I can't use it. How should I do without this theorem?

3) I have added the definitions used in my textbook...
 
#3 is shown more easily by the epsilon-delta definition of continuity. Have you studied that definition?

#2 is shown more easily using the convergent subsequence definition of compactness; have you studied that definition? Let \{x_n\} be a sequence in the preimage that converges to x_0 in the preimage. Do you understand why you can find such a sequence in the preimage?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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