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I know $\sin x$ is uniformly continuous and it was asked before (Prove $\sin x$ is uniformly continuous on $\mathbb R$). My question is related to this answer: https://math.stackexchange.com/a/219463/105885.

Uniform continuity of $\sin x$, if we'll take $x=\pi x +\frac {\pi} 2$ and $y=\pi x$, then $|x-y|={\pi \over 2}$ but similaly to the linked answer: $|f(x)-f(y)| = |1|\ge\epsilon = 1$.

Edit: With an arbitrarily small delta: $x=\pi n +\frac 1 n, \ y=\pi n$ then $|x-y|=\frac 1 n$ So delta can be arbitrarily small, $|f(x)−f(y)|=|\sin(\frac 1 n)|$ which for an infinite amount of $n$ can be $|\sin(\frac 1 n)|≥\frac 1 2$ (Since the function furiously alternate between $1$ and $-1$ as $n\to 0$).

Why this is wrong in showing that $\sin x$ isn't uniformly continuous?

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GinKin
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Because $|x-y|=\pi/2$ is not arbitrarily small.

Vladimir
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  • Can you explain a bit more ? Why does it matter if $\delta$ is arbitrarily small ? – GinKin Jun 21 '14 at 13:27
  • A function $f(x)$ is NOT uniformly continuous if, for some fixed $\epsilon>0$ and for ARBITRARILY SMALL $\delta$ there exist $x,y$ such that $|x-y|<\delta$ and $|f(x)-f(y)|>\epsilon$. Your $x$ and $y$ satisfy the second condition but not the first: $|x-y|=\pi/2$ and you cannot make it smaller. – Vladimir Jun 21 '14 at 13:30
  • Ok. What if in my example $x=\pi n +\frac 1 n$, then $|x-y|=\frac 1 n$ So delta can be arbitrarily small, $|f(x)-f(y)| = |\sin(\frac 1 n)|$ which for some $n$ can be $|\sin(\frac 1 n)|\ge 1$ – GinKin Jun 21 '14 at 13:55
  • $\sin\frac1n<\frac1n$ so it cannot be $\ge1$ for large $n$. – Vladimir Jun 21 '14 at 15:34