Theorem: Suppose that a sequence S of real numbers has a subsequence that converges to a real number a. Then S converges to a.

I know this is true as an if and only if statement, but I need a counter example to show that just one converging subsequence isn't enough.

Here are two sequences I'm considering: {1,-1,1,-1,1,-1...).
Another is {0, 1,1/2, 1/3, ..., 1/n}.

Our definition of a subsequence is: Let S={(n,an(subscript)}be a sequence of real numbers. A subsequence of S is a sequence T formed as follows:
1) Let theta be a function with domain the counting numbers and range contained in the counting numbers, and such that theta(n)>theta(m) whenever n>m.
2) T ={(m,a(theta(me)(subscript).
To specify a subsequence, you must specify the function theta.

Here's where I need help:
1) How do I develop a subsequence from one of the sequences I have. (If you have a simpler sequence, please suggust using it). What is the domain, range, and the function theta?

2) How do I show convergence? I know that I pick an r, but then what? I need to show how both S and T converge and that they don't converge to the same point.

Thanks!

Solution Preview

I will go ahead with your sequences.
First, we should understand that subsequence is a part of the given sequence. For the sequence you have considered above, we can list out few subsequences as {1, 1, 1, 1, 1, .....} or {-1, -1, -1, -1, -1, .....} or {1, -1, 1, 1, -1, 1, 1, 1, -1, 1, 1, 1, 1, -1.....}. A subsequence must be defined on its own but its elements must be from the parent sequence ...

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