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I would need some help with subsequences of a certain sequence.Let's $X$ be some space,and un a sequence of $X$.Let's suppose that $X$ is sequentially compact.Then let's $y_n$ is a subsequence of $u_n$,it converges.Now,$(u_n)-(y_n)$ is another sequence of $X$,then it has another subsequence an that converges,and then $(u_n)-(y_n)-(a_n)$ is a sequence...

My question is,how can we prove that this 'algorithm' has only a finite number of steps? (ie,there exists only a finite number of subsequences that can be constructed in this way) (I think it is true,but I can't think of a proof)

Thanks for your time

EDIT:I do not what an 'algorithm' (even though I said it,sorry for confusion),I really need to prove that there's only a finite number of sequences that can be constructed in this manner. Also,($u_n$)-($y_n$) means ($u_n$)/($y_n$)

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    Let $X = [0,1]$, and let $(u_n)$ be an enumeration of the rational numbers in $X$. Then no finite number of convergent subsequences ever exhausts $(u_n)$. – 2017-01-15
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    Thanks for your reply.It depends on how you define ($u_n$),but ($u_n$) itself is a subsequence converging to 1 (if you enumerate in a monotone way),no ? – 2017-01-15
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    You can't enumerate the rationals in $[0,1]$ in a monotone way. Every point in $[0,1]$ is an accumulation point of $(u_n)$, and if you take finitely many convergent subsequences, every $x\in [0,1]$ that is not one of the limits of these subsequences has a neighbourhood $(x-\varepsilon, x+\varepsilon)$ containing none of these limits. Then $(x-\varepsilon/2, x+\varepsilon/2)$ contains only finitely many terms of each subsequence, but it contains infinitely many terms of $(u_n)$. – 2017-01-15

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There are various wrong things in your question:

  1. As $X$ is sequentially compact, we can only infer that there exists a convergent subsequence, not that any subsequence converges.
  2. It makes no sense to speak of the difference of two sequences in a general topological space.

Even if you resolved those two problems, your "algorithm" would not be well-defined, as you don't give an algorithmic way to choose the subsequence. And even if you did find an algorithm taking an arbitrary sequence in a sequentially compact (say) topological vector space (which I don't think has any possibility to exist), then I'm pretty sure that it would be easy to construct a sequence for which the "algorithm" in your question would not converge.

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    Thanks for your answer,what I mean by the difference,is really the set containing all values of $u_n$ without the set containing all values of $y_n$. – 2017-01-15