Analysis: Prove that this function is not uniformly continuous on the interval $(0,1)$

Question

Prove that the function defined by

$$f(x)=\sin\left(\frac{1}{x}\right)$$

is not uniformly continuous on the interval $(0,1)$.

Answer 1

We need to show:

$$(\exists \epsilon >0) \space \space (\forall \delta >0) \space \space (\exists \space x, y \in \mathbb{R})$$ with $$\left |\space x-y\space \right|<\delta,\space\space\space\space \left|\space f(x)-f(y) \space\right|\geq\epsilon$$

Let $\epsilon=1$. Let $\delta>0$ be arbitrary. By the corollary of the Archimedean property $\exists n_0 \in \mathbb{N}$ such that $\frac{1}{n_0}<2\pi\delta$. Let $x_n = \frac{1}{2n\pi}$ and $y_n = \frac{1}{(2n+1/2)\pi}$.$\space$Then for every $n \in \mathbb{N}, n>n_0$ we have

$$\left|\space x_n-y_n\space \right|=\left|\space \frac{1}{2n\pi}-\frac{2}{(4n+1)\pi}\space \right|= \frac{1}{2n\pi(4n+1)}<\frac{1}{2n\pi}<\frac{1}{2n_0\pi}<\delta$$ but $$\left|\space f(x_n)-f(y_n)\space \right|=\left| \space sin(2n\pi)-sin((2n+\frac{1}{2})\pi) \space \right|=1\geq\epsilon.$$

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Answer 2

A uniformly continuous function sends cofinal sequences to cofinal sequences, that is, if $|x_n-y_n|\to 0$ and $f$ is uniformly continuous, then $|f(x_n)-f(y_n)|\to 0$. You should be able to prove this using the definition of uniformly continuous function. This doesn't happen with $\sin(x^{-1})$ over $(0,1)$, for the sequences defined by $1/x_n=2\pi n$ and $1/y_n = 2\pi n +1/2$ are cofinal in $(0,1)$ yet their image sequences are not.