 Solution

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe same $\delta$ must work at every point of $I$.

Every uniformly continuous function on $(0,1)$ is bounded. Choose the tolerance $\delta$ for $\epsilon =1$ and an integer $N$ with $1/N<\delta$. The finitely many centers $c_j=(j+1/2)/N$, $0\le j<N$, cover the interval by cells of radius $1/(2N)$. For each $x$, one center satisfies $|f(x)-f(c_j)|<1$. Thus $|f(x)|\le 1+{\max }_j|f(c_j)|$. This is the uniformly continuous function on a totally bounded set is bounded argument; closedness of the interval is unnecessary.

The reciprocal function is not uniformly continuous on $(0,1)$. It is unbounded, already contradicting the preceding result. Directly, $x_n=1/(n+1)$ and $y_n=1/(n+2)$ have distance tending to zero while $|1/x_n-1/y_n|=1$.

The oscillating function $\sin (1/x)$ is not uniformly continuous either, despite being bounded. TakeTheir distance tends to zero, but their function values are $1$ and $-1$. The fixed output gap of two contradicts the uniform-continuity condition, for example with $\epsilon =1$.