# I want to prove that the following set is closed

Let $A\subseteq R$ be a compact set and $B\subseteq R$ closed. Then $S=\{b\sin a;b\in B,a\in A\}$ is closed.

What I have done is to consider the continuous function $$f:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$$ defined by $f(x,y)=x\sin y.\;$ Then $\;S=f(B\times A).\;$ If $\,f\,$ is closed, then $\,S\,$ is closed. Is $\,f\,$ closed? ($\mathbb{R}$ is the real numbers)

Thanks!

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If $R$ stands for the real numbers, use \mathbb R or \Bbb R instead aliakbar. Nice question. + –  B. S. Jan 9 '13 at 10:23
@BabakSorouh Nice suggestion; I edited accordingly. –  amWhy Jan 9 '13 at 14:11
@amWhy: Thanks for doing that. :) –  B. S. Jan 9 '13 at 14:23

I am not sure if my example is wrong, or the question is inaccurate.

Consider $B=\{1,2,3,4,\ldots\}$, $A=\{0\}\cup\{\sin^{-1} 1, \sin^{-1} (1/2), \sin^{-1} (1/3),\ldots\}$

For example, $\sin^{-1} 1 \approx 1.5707, \sin^{-1} 1/2 \approx 0.5236, \sin^{-1} 1/3 \approx 0.3398$, etc.

$A$ is compact, $B$ is closed, and $S = f(B\times A) =\{0\}\cup \mathbb{Q}^+$, the non-negative rational numbers, which is not closed.

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I want know why the set $S$ is closed –  aliakbar Jan 9 '13 at 14:34
@aliakbar: this example (which looks fine to me) shows that $S$ is not closed in general, so it's going to be rather hard to show that it is closed, don't you think? –  Marek Jan 9 '13 at 15:00
@Marek Thank you very much –  aliakbar Jan 9 '13 at 15:08