Cauchy sequence with a convergent subsequence

[kingwinner]

Homework Statement

Theorem: In a metric space X, if (x_n) is a Cauchy sequence with a subsequence (x_{n_k}) such that x_{n_k} -> a, then x_n->a.

Homework Equations

N/A

The Attempt at a Solution

1) According to this theorem, if we can show that ONE subsequence of x_n converges to a, is that enough? Or do we need to show that EVERY subseuqence of x_n converges to a in order to claim that x_n->a?

2) How can we prove the theorem?
By definition, x_n is Cauchy iff for all ε>0, there exists N s.t. if n,m≥N, then d(a_n,a_m)<ε.
By definition, x_n->a iff for all ε>0, there exists M s.t. if n≥M, then d(x_n,a)<ε.
I know the definitions, but I don't see how to link these all together to prove the theorem...

Any help is greatly appreciated!

 

[rasmhop]

1) According to this theorem, if we can show that ONE subsequence of x_n converges to a, is that enough? Or do we need to show that EVERY subseuqence of x_n converges to a in order to claim that x_n->a?

You can only assume it for one subsequence. Otherwise you could just take the subsequence n_k=k and be done.

The general idea of this proof is that for a_nk to converge to a means that for large enough N for all k>N a_nk is close to a. Now the issue here is that a_nk doesn't include all elements of a_n, but to say that a_n is Cauchy means precisely that for large enough indices the elements are close to each other.

To actually perform the proof let us choose some $\epsilon > 0$. Note that since $a_{n_k} \to a$ there exists some K>0 such that for all k>K we have $d(a_{n_k},a) < \epsilon$. Now since $(a_n)$ is Cauchy there exists some N>0 such that if n,m>N, then $d(a_n,a_m) < \epsilon$. This is the necessary setup.

There exists some k' such that $n_{k'} > N$ and $k' > K$. By the Cauchy condition if $n>N$, then,
$$d(a_{n_{k'}},a_n) < \epsilon$$
Similarily by the convergence we have,
$$d(a_{n_{k'}},a}) < \epsilon$$
Add these and use the triangle inequality to get,
$$d(a_n,a) < 2\epsilon$$
Since $\epsilon$ was arbitrary this shows that $(a_n)$ converges to a.

 

[kingwinner]

You can only assume it for one subsequence. Otherwise you could just take the subsequence n_k=k and be done.

The general idea of this proof is that for a_nk to converge to a means that for large enough N for all k>N a_nk is close to a. Now the issue here is that a_nk doesn't include all elements of a_n, but to say that a_n is Cauchy means precisely that for large enough indices the elements are close to each other.

To actually perform the proof let us choose some $\epsilon > 0$. Note that since $a_{n_k} \to a$ there exists some K>0 such that for all k>K we have $d(a_{n_k},a) < \epsilon$. Now since $(a_n)$ is Cauchy there exists some N>0 such that if n,m>N, then $d(a_n,a_m) < \epsilon$. This is the necessary setup.

There exists some k' such that $n_{k'} > N$ and $k' > K$. By the Cauchy condition if $n>N$, then,
$$d(a_{n_{k'}},a_n) < \epsilon$$
Similarily by the convergence we have,
$$d(a_{n_{k'}},a}) < \epsilon$$
Add these and use the triangle inequality to get,
$$d(a_n,a) < 2\epsilon$$
Since $\epsilon$ was arbitrary this shows that $(a_n)$ converges to a.

There exists some k' such that $n_{k'} > N$ and $k' > K$ <---can you please explain this part?

Thank you!

 

[rasmhop]

There exists some k' such that $n_{k'} > N$ and $k' > K$ <---can you please explain this part?

Thank you!

Well $1 \leq n_1 < n_2 < \cdots$ so $n_i \geq i$ (n_1 must be at least 1, n_2 must be at least one more, n_3 must be at least one more than n_2, etc.). Thus $n_{N+1} > N$ so simply let $k' = \max(N+1,K+1)$, then
$$n_{k'} \geq n_{N+1} > N \qquad k' \geq K+1 > K$$
Basically what this says is the sequence (n_k) will eventually become greater than N (it's unbounded) and there exists an integer larger than K.