Sequence is convergent if it has a convergent subsequence

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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 [itex]\exists[/itex]N such that jn>N implies |xjn-x|<[itex]\epsilon[/itex]
It follows that n>jn>N implies |xn-x|<[itex]\epsilon[/itex]

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.
 

Answers and Replies

  • #2
jbunniii
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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 [itex]\exists[/itex]N such that jn>N implies |xjn-x|<[itex]\epsilon[/itex]
It follows that n>jn>N implies |xn-x|<[itex]\epsilon[/itex]
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 [itex]x_n[/itex] is an increasing sequence. This would not be true for an arbitrary sequence.
 
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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  [itex]\epsilon[/itex]> 0 there is an N such that n > N implies |xn - x| < [itex]\epsilon[/itex] . Now
consider a subsequence (xjn). Then since jn [itex]\geq[/itex] n > N we have that for any n > N,
|xjn - x| < [itex]\epsilon[/itex]  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.
 
  • #4
jbunniii
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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 [itex]\epsilon[/itex]> 0 there is an N such that n > N implies |xn - x| < [itex]\epsilon[/itex] . Now
consider a subsequence (xjn). Then since jn [itex]\geq[/itex] n > N we have that for any n > N,
|xjn - x| < [itex]\epsilon[/itex] 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 [itex]x_n[/itex] converges to [itex]x[/itex], then [itex]x_n[/itex] converges to [itex]x[/itex]. This is trivial, because [itex]x_n[/itex] is a subsequence of itself.

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

A general sequence can have some subsequences which converge, and others which do not. For example, let [itex]x_n = 0[/itex] if [itex]n[/itex] is even, and [itex]x_n = n[/itex] if [itex]n[/itex] 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.
 
  • #5
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Thanks it makes sense.
 

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