I am very confused at everything to do with subsequences. The author defines a subsequence as the following:

Given a sequence:

$(a_n) = (1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, ...)$

let $n_1 \lt n_2 \lt n_3 \lt n_4 \lt ...$ be an increasing sequence of natural numbers.

Then the sequence

$a_{n_1}, a_{n_2}, a_{n_3}, a_{n_4}, ...$ is called a subsequence of $(a_n)$ and is denoted by $(a_{n_j})$ with $j \in \mathbb{N}$.

All fine and dandy. A subsequence is a sequence composed of elements of another sequence in the same order, without repeats. It is basically "dropping" all indexes but the chosen $j's$.

The author now wishes for me to prove that

Subsequences of a convergent sequence converge to the same limit as the original sequence.

What I know is that is a convergent sequence is bounded, so intuitively this bound $M \gt 0$ will be the limit of $a_n$. I started a bit of a proof and immediately got lost. Lets say I have a sequence:

$\{1, 2, 3, 4, 5, 6, 7\}$.

Intuitively, the limit of this sequence is $7$. But if I take any subsequence:

$\{2, 4, 6\}$

It's limit is $6$, is it not? How can this subsequence's limit be $7$?

I don't want help with the proof as much as help clearing up this really bad misunderstanding of subsequences. I would prefer to struggle with the proof with the intuition so that I can fully grasp it.

My though was that since a sequence can have many upper bounds (for example, $7, 8, 9, 10, ...$ all are upper bounds for my example sequence) this would lead to the intuition. Another thought I had was subsequences in the question referred to the set of all subsequences of the convergent sequence, in which case we would be talking about something like a limit superior which would make more sense. However, I am hopelessly lost.

Can anyone explain to me the intuition that leads one to recognize a subsequence's limit is the limit of its "parent sequence"?

  • $\begingroup$ A sequence (and a subsequence) should be infinite in length. $\endgroup$ – Minus One-Twelfth Feb 21 '19 at 5:10

The issue is that all sequences are implicitly infinite in this context. Sometimes in my texts, what you write $(a_n)$ in your post, I would see written as $(a_n)_{n\in\Bbb N}$ in my work. We only drop the "$n\in\Bbb N$" bit when it's understood what is going on (that it's infinite).

That is,

$$(a_n)_{n\in\Bbb N} = (a_1, a_2, a_3, a_4, ... \text{and so on and so forth, forever neverending})$$

Of course, then, by convention, we have that the sequence $n_1,n_2,n_3,...$ (where each term is strictly greater than the last) which defines the indices of the subsequence is also infinite. That is to say, every sequence (in this context) is infinite, as are all subsequences.

It is in fact that infiniteness that allows us to conclude that each subsequence of a convergent sequence converges to the same limit as the sequence itself - as you have seen, it doesn't hold in the finite case.

Also, for what it's worth, this theorem is discussed on MSE here.

  • $\begingroup$ This makes so much more sense. So basically in a subsequence we are guaranteed to include the end point at some point? I had figured that from the ellipses in the definition, but the question confused me with its wording. $\endgroup$ – CL40 Feb 21 '19 at 5:33
  • $\begingroup$ There is no endpoint. It's sort of like how, in calculus, we have this notion of "the sequence of partial sums" when finding an infinite sum. There may not be a point in the sequence which is equal to the limit, but we closer and closer to that limit, effectively. For example, the sequence $(a_n) = (1/n) = (1,1/2,1/3,...)$ doesn't ever have a term equal to $0$, but $a_n \to 0$ as $n \to \infty$. The "..." - with nothing following it - is basically saying "there's infinitely many terms here following the pattern" (and of course, there is no "endpoint" in an infinitely long sequence). $\endgroup$ – Eevee Trainer Feb 21 '19 at 5:37
  • $\begingroup$ Ah yes, that's what I meant. Sorry it was informal. I'll go ahead and mark this answered now, thank you! $\endgroup$ – CL40 Feb 21 '19 at 5:39

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