Subsequences. Prove that (xn) is bounded. Prove that (xn) converges to x.
Let x be a real number, and suppose (xn) is a sequence such that every subsequence (xnk ) of (xn) has a subsequence (xnkj ) that converges to x. (a) Prove that (xn) is bounded. (b) Prove that (xn) converges to x.
I think I see what is going on in this exercise. But I don't see how to work backwards clearly. From sub subsequence back to the original sequence and have a rigorous proof.
Here is my attempt.
I started first by stating that lim Xnkj=x and Xnkj is bounded.
So for every ε > 0 exists N, such that n>N implies |xnkj-x|<ε for all n>N.
But I don't see how do I go backwards for proving that Xnk is bounded first.
Thanks for help I still need help with part b. Comments are below.
$\endgroup$ 11 Answer
$\begingroup$1) Suppose $(x_n)_n$ is not bounded. Then given $N > 0$ there exists $|x_N| > N$. But then we have a subsequence that tends to infinity contradicting that every subsequence converges to $x$.
2) $x_n \to x$ Suppose not. Then there exists $\epsilon > 0$ and $n_1, n_2, ...$ such that $|x_{n_k} - x| \geq \epsilon$ contradicting that every subsequence converges to $x$.
$\endgroup$ 3
