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Prove that the sequence of functions $f_n(x) = \sin(n x)$ have no pointwise convergent subsequence

I am confused. If I let $x = 1$, then we get $f_n(1) = \sin(n)$. By Bolzano Weistress, we have a convergent subsequence no?

Lemon
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1 Answers1

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A pointwise convergent subsequence would be a (sub)sequence of functions $f_i(x)$ such that $\lim_{i\to \infty} f_i(x)$ exists, for all $x$.

vadim123
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  • You wrote $\lim_{i \to \infty} f_i(x)$; isn't that just regular pointwise convergence? – Lemon Jun 16 '13 at 21:55
  • Correct. It's not the whole sequence $f_n(x)$ that converges pointwise, but a subsequence; i.e. choosing some functions but not others. – vadim123 Jun 16 '13 at 22:09
  • Right, and I chose $f_i(1)$, which shows there is a subsequence – Lemon Jun 16 '13 at 22:12
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    You need convergence of some functions, for all $x$. What you did was the convergence of some functions, for one $x$. – vadim123 Jun 16 '13 at 22:14