# How to prove sin(nx) has no pointwise convergent subsequence without prior knowledge of Lebesgue's Theory?

In Baby Rudin 7.20 example, the author mentions to prove that the function sequence $$f_n(x):=\sin(nx) \qquad0(\leq x\leq 2\pi)$$has no pointwise convergent subsequence would be troublesome without Lebesgue's Theorem.

Is there a proof that doesn't refer to Lebesgue's Theorem, and only requires the knowledge introduced in the first 7 chapter in Rudin?

Here's an argument my officemate and I came up with that should work (while avoiding Lebesgue theory). Given a subsequence $$\sin(n_kx)$$, it constructs a sub-subsequence $$\sin(n_{k_m}x)$$ and a point $$y$$ such that $$\sin(n_{k_m}y)$$ fails to converge.
Suppose a subsequence $$\sin(n_kx)$$ is given. Then, we can find some closed interval $$I_1 \subseteq [0, 2\pi]$$ such that $$\sin(n_1x)$$ maps $$I_1$$ into $$[1/2, 1]$$. Also, set $$n_{k_1} = n_1$$.
Having chosen $$I_1$$, observe that for $$k > k_1$$ large enough, $$\sin(n_k x)$$ maps $$I_1$$ to $$[-1,1]$$ surjectively. Choose $$k_2$$ to be the least $$k>k_1$$ with this property, and let $$I_2 \subset I_1$$ be an interval that $$\sin(n_{k_2}x)$$ maps to $$[-1, -1/2]$$.
In general, suppose we have constructed the index $$n_{k_m}$$ and interval $$I_m$$. Then, we let $$k_{m+1}$$ be the least integer greater that $$k_m$$ such that $$\sin(n_{k_{m+1}}x)$$ maps $$I_m$$ surjectively onto $$[-1, 1]$$. Then, if $$m$$ is odd, we choose a subinterval $$I_{m+1} \subset I_m$$ such that $$\sin(n_{k_{m+1}}x)$$ maps $$I_{m+1}$$ into $$[-1, -1/2]$$, while if $$m$$ is even, we choose an $$I_{m+1}$$ which $$\sin(n_{k_{m+1}}x$$ maps into $$[1/2,1]$$.
Now, by the nested intervals theorem, $$\bigcap_{m=1}^\infty I_m$$ is nonempty, so it contains some point $$y$$. But, by construction, $$\sin(n_k y)$$ is both at least $$1/2$$ and at most $$-1/2$$ infinitely often, so the sequence of functions cannot converge at $$y$$.
• Nice proof. BTW at the beginning I think you meant to write $n_{k_1} = n_1$. – Michael Feb 19 '19 at 9:39