Sequence satisfying |x(n) - x(n - 1)| --> 0, can you help me prove this?

Question

Suppose such sequence is in a compact set of R^n. Show the set of its subsequential limits is connected.

Thank you.

Answer 1

I think the 2 previous posters didn't get the precise meaning of what is being asked. This sequence doesn't need to converge, it's not necessarily Cauchy. And actually, this conclusion is not restricted to Euclidean spaces, it's true in any metric space.


Let A be a compact set of R^n that contains x(n) and let S be the set of the subsequential limits of x(n). We know the set of subsequential limits of any sequence in a metric space is closed. So, S is a closed subset of R^n and is contained in the closure of A. Since A is compact, A is closed and is its own closure, so that S is contained in A. So, S is compact, because closed subsets of compact sets are compact.

Suppose S is disconnected and let S1 and S2 form a disconnection of S. Then, S1 and S2 are not empty, disjoint and closed with respect to S. Since S is compact, so are S1 and S2, and, since S1 and S2 are disjoint, it follows the distance between them is d(S1, S2) = infimum {|s 1 - s2| | s1 is in S1, s2 is in S2} > 0. So, there are open sets O1 containing S1 and O2 containing S2 with d(O1, O2) > 0.

Let K be the complement of O1 U O2 with respect to A, that is, K = A - (O1 U O2). Then, K is a closed subset of the compact A, which implies K is compact. If x(n) contained infinitely many terms in K, then there'd be a subsequence of x(n) in K. So, by compactness of K, this subsequence would have a subsequence converging to an element of K, contradicting the fact that the subsequential limits of x(n) are in S contained in O1 U O2. Therefore, x(n) has a tail in O1 U O2 and we can assume, without loss of generality, that x(n) is a sequence in O1 U O2.

Since |x(n) - x(n-1)| --> 0, there's p such that n > p => |x(n) - x(n-1)| < d(S1, S2). Pick an m > p. Them x(m) is in either in O1 or O2, say O1, WLOG. Suppose that, for some n >=m, x(n) is in O1. Then |x(n+1) - x(n)| < d(S1, S2), which implies x(n+1) is in O1. By induction, we conclude that, except for a finite number of terms, x(n) is in O1, which implies all of its subsequential limits are in the closure of O1. Then, S1 is a subset of this closure, and since d(O1, O2) >0, the closures of O1 and O2 don't intersect. It follows no subsequential limit of x(n) is in O2, so none is in S2 and S2 is, therefore, empty. But this contradicts the fact that S1 and S2 form a disconnection of S.

It follows S is connected and we are done.

The same proof is valid in any metric space. All you have to do is replace | . | by the distance function of the metric space.

Steiner
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Answer 2

with the aid of fact the shrink as x->0 f(x) isn't comparable to L, there exists ?_0 >0 such that for each ? > 0 there exists x such that 0 < |x - 0| < ? yet |f(x) - L| ? ?_0. Now evaluate the series of delta values ?_n = a million/n. with the aid of the above, we are in a position to construct a chain {x_n} interior here way: for each n, choose for x_n such that 0 < |x_n - 0| < ?_n and |f(x_n) - L| ? ?_0. it relatively is, lim(n->?) f(x_n) ? L. It purely keeps to be to instruct that x_n -> 0. be conscious that for any ? > 0, we are in a position to compliment N such that N ? a million/?. Then for each n ? N we've |x_n - 0| < a million/n ? ?. subsequently x_n -> 0. So we've built a chain {x_n} convergent to 0 and such that f(x_n) would not converge to L.

Answer 3

Y1= X
Y2= X-1
Y3= Y1-Y2
Y4= |Y3|
graph on a calculator
or even simpler, def. of absolute value: positive distance from zero, so it's always greater than or equal to zero

broken down further:
|x(n)- x(n-1)|
|xn- xn + x|
|x|
just make a chart showing input as -1, -2, -3...
with outputs 1, 2, 3...