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I've been studying real analysis recently from "Introduction of real analysis" by Robert G. Bartle. There is a theorem in a book that goes like,

If a sequence X=(xn) of real numbers converges to a real number x then any subsequence X'=(xnk) of X also converges to x.

I get the intuition of what this theorem suggest but what if the sequence we have is finite and subsequence of this sequence don't include the last term?

Chris Culter
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Harshit Halwan
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  • What do you mean by a finite sequence? A sequence of real numbers is a function $x:\mathbb{N}\to\mathbb{R}$, so there are always infinitely many terms. – carmichael561 May 27 '17 at 03:58
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    What does it mean to say a finite sequence "converges"? Said differently, when talking about convergent sequences, it's assumed, unlesss specified otherwise, that a sequence $(a_n)$ is an infinite sequence: $$(a_n) = (a_1,a_2,a_3,...)$$ – quasi May 27 '17 at 03:58
  • @carmichael561 That's more of a convention. You can also define a sequence as a mapping from a convex subset of $\mathbb{N}$ to $\mathbb{R}$. – MathematicsStudent1122 May 27 '17 at 04:03
  • @MathematicsStudent1122: You can define anything you want however you please, but the definition I gave is the standard one. – carmichael561 May 27 '17 at 04:05
  • @carmichael561 According to whom? Googling 'finite sequence' gives quite a bunch of hits, and many of them quite reliable. – MathematicsStudent1122 May 27 '17 at 04:06
  • @carmichael561 Also, we can index sequences however we want. If the index starts at $n=5$, say, it's still a sequence, is it not? (ie., ${a_n}_{n \geq 5}$) – MathematicsStudent1122 May 27 '17 at 04:08
  • @MathematicsStudent1122 thans for clarifying, that's what I'm trying to ask. Let our sequence have 50 terms and the subsequence is the every term at odd place then the subsequence will not include the 50th term. But the theorem here doesn't hint for any such situations. Please don't mind my language I'm a bit too slow in writing maths here, so i wrote my intuition. – Harshit Halwan May 27 '17 at 04:13

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Limits of finite sequences aren't terribly interesting. That said, finite sequences do qualify as nets, and the usual definition of the limit of a net gives you that the limit is the last element of the sequence. See also the answers to Can a sequence be called convergent/divergent if it has finite number of terms? and linked question(s).

Under these definitions, the sequence $\mathbf X:(1,2,3,4)\mapsto(6,7,8,9)$ converges to $\mathbf x=9$, while the sequence $\mathbf X' :(1,2,3)\mapsto(6,7,8)$ converges to $8$. So IF we're willing to call this $\mathbf X'$ a subsequence of $\mathbf X$, then you're right, a subsequence of a finite sequence need not converge to the same limit!

There is no contradiction with the theorem stated in Bartle's textbook, because (I assume) that book only considers infinite sequences.

All that said, the theorem can be generalized to arbitrary nets, including finite sequences! There are two slightly different ways to do it.

(1) If a net converges to a point $\mathbf x$, then any cofinal subnet also converges to $\mathbf x$. This theorem doesn't apply to our example, as $\mathbf X'$ is not a subnet of $\mathbf X$, because its index set $(1,2,3)$ is not cofinal in $(1,2,3,4)$, because it doesn't include the maximum index $4$.

(2) Even better, if a net converges to a point $\mathbf x$, then any reindexed subnet also converges to $\mathbf x$. This is more like what Bartle is doing, considering a subsequence of $(x_n)$ to be a reindexing like $(x_{n_k})$. If you go back to the definition of a subsequence, you'll see a condition like "the reindexing sequence $(n_k)$ must eventually be $\geq M$ for all $M$". In the context of a finite index set for $(x_n)$, that means that $n_k$ must include the maximum index. Again, in our case that's $4$.

Given (1) and (2), we may find it more convenient to define subsequences of finite sequences such that $\mathbf X'$ is not a subsequence of $\mathbf X$ at all!

Chris Culter
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