# Sequence is convergent if it has a convergent subsequence

## Homework Statement

Show that an increasing sequence is convergent if it has a convergent subsequence.

## The Attempt at a Solution

Suppose xjn is a subsequence of xn and xjn→x.

Therefore $\exists$N such that jn>N implies |xjn-x|<$\epsilon$
It follows that n>jn>N implies |xn-x|<$\epsilon$

Therefore xn→x

The solution that I've been given is much more complicated I'm just wondering whether my simpler solution is also correct.

jbunniii
Homework Helper
Gold Member

## Homework Statement

Show that an increasing sequence is convergent if it has a convergent subsequence.

## The Attempt at a Solution

Suppose xjn is a subsequence of xn and xjn→x.

Therefore $\exists$N such that jn>N implies |xjn-x|<$\epsilon$
It follows that n>jn>N implies |xn-x|<$\epsilon$
It's true, but WHY does it follow? This is the key part of the proof, so you need to be explicit. You need to use the fact that $x_n$ is an increasing sequence. This would not be true for an arbitrary sequence.

The original question was an 'if and only if question' which means the reverse also had to be proved ie:

Show that if xn → x that any subsequence of (xn) also converges to x
My solution which is simliar to the answer given is

If xn → x then given any  $\epsilon$> 0 there is an N such that n > N implies |xn - x| < $\epsilon$ . Now
consider a subsequence (xjn). Then since jn $\geq$ n > N we have that for any n > N,
|xjn - x| < $\epsilon$  and so we conclude that the subsequence has the same limit

I believe this correct and this raises the question why does logic work in the one direction but no the other.

jbunniii
Homework Helper
Gold Member
The original question was an 'if and only if question' which means the reverse also had to be proved ie:

Show that if xn → x that any subsequence of (xn) also converges to x
My solution which is simliar to the answer given is

If xn → x then given any $\epsilon$> 0 there is an N such that n > N implies |xn - x| < $\epsilon$ . Now
consider a subsequence (xjn). Then since jn $\geq$ n > N we have that for any n > N,
|xjn - x| < $\epsilon$ and so we conclude that the subsequence has the same limit

I believe this correct and this raises the question why does logic work in the one direction but no the other.

It does work in both directions. If EVERY subsequence of $x_n$ converges to $x$, then $x_n$ converges to $x$. This is trivial, because $x_n$ is a subsequence of itself.

But the hypothesis in the first part is weaker: $x_n$ has *a* convergent subsequence. Without the additional assumption that $x_n$ is increasing, this would not be enough to conclude that $x_n$ converges.

A general sequence can have some subsequences which converge, and others which do not. For example, let $x_n = 0$ if $n$ is even, and $x_n = n$ if $n$ is odd. The subsequence consisting of even indices converges, and the subsequence consisting of odd indices diverges. And of course the sequence itself does not converge.

Thanks it makes sense.