Takeover Season
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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.

Attachment 930730

Attachment 930732

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!
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Gregorius
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(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,...
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Gregorius
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(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 \epsilon.

If you choose \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 \epsilon does it obey the rule? If you choose \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 \epsilon.
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(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.
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(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 \epsilon.

If you choose \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 \epsilon does it obey the rule? If you choose \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 \epsilon.
Hi, thanks for your reply.
Yes, so if I choose any \epsilon &lt; 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 \epsilon \geq 1, then this proof doesn't work as then d(a_n, a_m) &lt; \epsilon \not\Rightarrow d(a_n, a_m) = 0 as we could have d(a_n, a_m) = 0 or =1. So, how do I show the sequence is convergent for \epsilon \geq 1. Maybe it is because I am confused when you said "If you choose \epsilon \ge 1 then any sequence obeys the rule".

Thank you
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Gregorius
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(Original post by Takeover Season)
Hi, thanks for your reply.
Yes, so if I choose any \epsilon &lt; 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 \epsilon \geq 1, then this proof doesn't work as then d(a_n, a_m) &lt; \epsilon \not\Rightarrow d(a_n, a_m) = 0 as we could have d(a_n, a_m) = 0 or =1. So, how do I show the sequence is convergent for \epsilon \geq 1. Maybe it is because I am confused when you said "If you choose \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 \epsilon. So a sequence obeying that condition for any particular value of \epsilon may or may not be a Cauchy sequence.
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(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 \epsilon. So a sequence obeying that condition for any particular value of \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 \epsilon &gt; 0, so it holds for \epsilon = 1/2 as well, and so if we can show the sequence is eventually constant for \epsilon = 1/2 (or just any one choice of \epsilon, then we are done?
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Gregorius
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(Original post by Takeover Season)
Oh right, so the sequence in the question is a Cauchy sequence. Its limiting condition holds for all \epsilon &gt; 0, so it holds for \epsilon = 1/2 as well, and so if we can show the sequence is eventually constant for \epsilon = 1/2 (or just any one choice of \epsilon, then we are done?
Yes.
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(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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