subsequences?

Question

suppose that every subsequence of X = x-n
have asubsequence that converges to 0 .
and show that lim X=0

Answer 1

This is a nice little problem. I've never seen it before, but it is typical of a class of relatively simple proofs.

The key, of course, is trying to establish that if the proposition is false, we have a contradiction.

In this case, that means that if there is a sequence that does not converge to 0, then there must be at least one subsequence that does not have a subsequence that converges to 0.

So how can we know that no subsequence of a sequence converges to 0? Well, if |Xn| > epsilon for all n and for epsilon > 0, then no subsequence of Xn can converge to 0.

Taking it from the top.

Let Xn be a sequence such that every subsequence of Xn has a subsequence that converges to 0 and such that Xn itself does not converge to 0.

To converge to 0 means that for any epsilon e, there exists an N such that n > N ==> |Xn| < e

http://en.wikipedia.org/wiki/Limit_of_a_sequence

So, if Xn does not converge to 0, it follows that there exists an epsilon e, such that no N exists that guarantees n > N ==> |Xn| < e

That means that for this epsilon, for each N, there is at least one k such that k > N and |Xk| >= e.

So if we create a subsequence of Xn from these Xk's, we are done. No subsequence of these Xk's can converge to 0.

There are many ways to create this subsequence.

We have a special epsilon e (one that doesn't have an N)

Let k(1) be the first n such that |Xn| >= e
Then let k(i+1) the first n > (k(i) + 1) such that |Xn| >= e

So the k(i)'s form an infinite subseqence for the integers so Xk(i) is a legitimate subsequence of Xn and has the property we want that |Xk(i)| >= e

QED.

Answer 2

outline f(x) = 0,a million/(0,a million+x) for x > 0,a million. the form of f is the open era (d192e0c4ad64a9c35fe32972477e4cd8d192e0c4ad64a9c35fe32972477e4cd8d192e0c4ad64a9c35fe32972477e4cd8). f'(x) = -- 0,a million / (0,a million+x)^2 so f is precisely lowering . considering that f(0,a million) = 0,a million and f(0,a million) = 0,a million/2d192e0c4ad64a9c35fe32972477e4cd8 f maps (d192e0c4ad64a9c35fe32972477e4cd8d192e0c4ad64a9c35fe32972477e4cd8d192e0c4ad64a9c35fe32972477e4cd8) = J_d192e0c4ad64a9c35fe32972477e4cd8 0,a million-0,a million onto J_2 = (0,a million/2d192e0c4ad64a9c35fe32972477e4cd8 0,a million)0,a million f maps J_2 to J_3 = (0,a million/2d192e0c4ad64a9c35fe32972477e4cd8 2/3)0,a million etc. be conscious that |f'(x)| < 4/9 < 0,a million/2 if x > 0,a million/2. consequently if we outline periods J_(n+0,a million) = f(J_n)0,a million J_(n+0,a million) is a subinterval whose length is decrease than 0,a million/2 the scale of J_n. If m & n are > N then x_m & x_n are in J_Nd192e0c4ad64a9c35fe32972477e4cd8 so |x_m -- x_n| < 0,a million/2^(N-0,a million) consequently {x_n: n > 0,a million} is a Cauchy sequenced192e0c4ad64a9c35fe32972477e4cd8 so converges to a shrink L in [d192e0c4ad64a9c35fe32972477e4cd8d192e0c4ad64a9c35fe32972477e4cd8d192e0c4ad64a9c35fe32972477e4cd8]. for sure L = f(L) = 0,a million/(0,a million + L)0,a million so L^2 +L -- 0,a million = 0,a million. The advantageous root of this quadratic is ( -- 0,a million + sqrt(5))/2 . ------------