The Student Room Group

Limit / Convergence

Hi there,

1. What is the intuition behind "Every subsequence of a divergent real sequence is divergent" being false.

I get it why it is true for convergence sequences. For example, if a sequence is convergent, then eventually after a certain term, all of the sequence's term converge to that limit L. So, since any subsequence's terms are also terms of the sequence, eventually there will be a certain term in the subsequence for which terms after it will also converge to that same limit L, but the 'certain terms' may be different as the subsequence's terms are a subset of the sequence's terms, and not necessarily all of them.

I was wondering what intuition helps me to think of "Not all subsequences of a divergent real sequence are also divergent". I get the examples given, very easy ones and straightforward. But, I was thinking of an intuitive idea, if anyone has one.

2. The following question is regarding the screenshots below. I'll tell you that the discrete metric is a metric such that the distance between any two same terms is 0 and if the two terms are different, the distance between them is 1.

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In this proof, we used used ϵ = 1/2, I can see that it will work for any 0 < ϵ < 1. Earlier on in my course, we proved that "Let X be a nonempty set equipped with the discrete metric. Show that a sequence (an)n∈N is convergent if and only if it is eventually a constant sequence (that is, there is a c X and an N N such that for all n>N, an = c)."

So, I am just wondering that for the definition of convergence, don't we have to show it '"for all ϵ > 0". So, if we chose e.g. any ϵ >= 1, then what happens? This proof doesn't work if we choose any ϵ >= 1. It seems like we just use an ϵ for which we get a constant sequence and so the sequence must be convergent.

So, can I get it clear that, if we were showing convergence using the definition, we'd have to show that "For all ϵ > 0, there exists a ..... ..... such that .... d(an,L) < ϵ" whereas if you are showing convergence using the fact that the sequence is eventually constant, then you only have to show it for one particular ϵ which makes the sequence eventually constant and hence the sequence is convergent. Is this right?

Thank you!
Original post by Takeover Season
Hi there,

1. What is the intuition behind "Every subsequence of a divergent real sequence is divergent" being false.



Take an arbitrary convergent sequence, say 0,0,0,... and embed it as every other term in a divergent sequence, say 1,2,3,... to get 0,1,0,2,0,3,...
Original post by Takeover Season


So, I am just wondering that for the definition of convergence, don't we have to show it '"for all ϵ > 0". So, if we chose e.g. any ϵ >= 1, then what happens? This proof doesn't work if we choose any ϵ >= 1. It seems like we just use an ϵ for which we get a constant sequence and so the sequence must be convergent.


The logic of the proof is as follows:

To show completeness, you require that every Cauchy sequence converges.

To be a Cauchy sequence, you require a condition to hold for all choices of ϵ\epsilon.

If you choose ϵ\epsilon to be 1/2 (or any positive value less than one), you note that this implies that the sequence is eventually constant.

An eventually constant sequence is a convergent sequence.

Think of the condition to be a Cauchy sequence as a "challenge/response" condition. Given a sequence, and my choice of ϵ\epsilon does it obey the rule? If you choose ϵ1\epsilon \ge 1 then any sequence obeys the rule; but this is not enough, as the sequence has to obey the rule for any choice of ϵ\epsilon.
Original post by Gregorius
Take an arbitrary convergent sequence, say 0,0,0,... and embed it as every other term in a divergent sequence, say 1,2,3,... to get 0,1,0,2,0,3,...

Oh yes thank you. That makes a lot more sense, so if you combine any convergent sequence with a divergent sequence, it becomes divergent. But, as you've formed it using a convergent sequence, it will always have that as its convergent subsequence, so not all subsequences of real divergent sequences are also divergent. Perfect.
Original post by Gregorius
The logic of the proof is as follows:

To show completeness, you require that every Cauchy sequence converges.

To be a Cauchy sequence, you require a condition to hold for all choices of ϵ\epsilon.

If you choose ϵ\epsilon to be 1/2 (or any positive value less than one), you note that this implies that the sequence is eventually constant.

An eventually constant sequence is a convergent sequence.

Think of the condition to be a Cauchy sequence as a "challenge/response" condition. Given a sequence, and my choice of ϵ\epsilon does it obey the rule? If you choose ϵ1\epsilon \ge 1 then any sequence obeys the rule; but this is not enough, as the sequence has to obey the rule for any choice of ϵ\epsilon.

Hi, thanks for your reply.
Yes, so if I choose any ϵ<1\epsilon < 1, this proof follows through with the fact that the Cauchy sequence is eventually constant and hence convergent.

But, my confusion was that if I choose any ϵ1\epsilon \geq 1, then this proof doesn't work as then d(an,am)<ϵ⇏d(an,am)=0d(a_n, a_m) < \epsilon \not\Rightarrow d(a_n, a_m) = 0 as we could have d(an,am)=0d(a_n, a_m) = 0 or =1=1. So, how do I show the sequence is convergent for ϵ1\epsilon \geq 1. Maybe it is because I am confused when you said "If you choose ϵ1\epsilon \ge 1 then any sequence obeys the rule".

Thank you
Original post by Takeover Season
Hi, thanks for your reply.
Yes, so if I choose any ϵ<1\epsilon < 1, this proof follows through with the fact that the Cauchy sequence is eventually constant and hence convergent.

But, my confusion was that if I choose any ϵ1\epsilon \geq 1, then this proof doesn't work as then d(an,am)<ϵ⇏d(an,am)=0d(a_n, a_m) < \epsilon \not\Rightarrow d(a_n, a_m) = 0 as we could have d(an,am)=0d(a_n, a_m) = 0 or =1=1. So, how do I show the sequence is convergent for ϵ1\epsilon \geq 1. Maybe it is because I am confused when you said "If you choose ϵ1\epsilon \ge 1 then any sequence obeys the rule".

Thank you

The point is that we're only interested in the convergence of Cauchy sequences. A sequence is a Cauchy sequence only if it obeys that limiting condition for all possible value of ϵ\epsilon. So a sequence obeying that condition for any particular value of ϵ\epsilon may or may not be a Cauchy sequence.
Original post by Gregorius
The point is that we're only interested in the convergence of Cauchy sequences. A sequence is a Cauchy sequence only if it obeys that limiting condition for all possible value of ϵ\epsilon. So a sequence obeying that condition for any particular value of ϵ\epsilon may or may not be a Cauchy sequence.

Oh right, so the sequence in the question is a Cauchy sequence. Its limiting condition holds for all ϵ>0\epsilon > 0, so it holds for ϵ=1/2\epsilon = 1/2 as well, and so if we can show the sequence is eventually constant for ϵ=1/2\epsilon = 1/2 (or just any one choice of \epsilon, then we are done?
Original post by Takeover Season
Oh right, so the sequence in the question is a Cauchy sequence. Its limiting condition holds for all ϵ>0\epsilon > 0, so it holds for ϵ=1/2\epsilon = 1/2 as well, and so if we can show the sequence is eventually constant for ϵ=1/2\epsilon = 1/2 (or just any one choice of \epsilon
, then we are done?
Yes.
Original post by Gregorius
Yes.


Ah thank you, that makes sense. Earlier on, I was thinking that because when we show it using the limit definition, we need to show it holds for all epsilon too. But, here we can use a particular epsilon to show the Cauchy sequence is constant and hence must be convergent and we don't need to show it for all epsilon as in the limit way.

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