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Example 3.3.5(a):

The sequence does not converge, but we can extract a convergent subsequence.
Since | sin(x) | < 4, the sequence is clearly bounded above and below (the sequence is also, of course, bounded by 1).

Therefore, using the Bolzano-Weierstrass theorem, there exists a convergent subsequence.

However, it is nearly impossible to actually list this subsequence. The Bolzano-Weierstrass theorem does guaranty the existence of that subsequence, but says nothing about how to obtain it.

The original sequence { sin(j) }, incidentally, does not converge. The proof of this is not so easy, but if we assume that the second part of this example has been proved, it would be easy. Remember that the second part of this example states that given any number L with |L| < 1 there exists a subsequence of that converges to L. If that was true the original sequence can not converge, because otherwise all its subsequences would have to converge to the same limit.

Of course this proof is only valid if this - more complicated - statement can be proved.

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