TITLE: Is function $y=\tan x$ uniformly continuous in the open interval $(0,\pi/2)\;?$
QUESTION [5 upvotes]: Determine whether the function $y=\tan x$ is uniformly continuous in the open interval $(0,\pi/2)$.
I tried approaching it this way
Let $x,y  \in (0, \pi/2)$. Then
$$|f(x)-f(y)|=|\tan x-\tan y|={\left|{{\sin x\cos y-\cos x\sin y}\over \cos x\cos y}\right|}\le|\sin(x-y)|\le|x-y|$$
Selecting $\delta=\epsilon$ we have that the given function is uniformly continuous.
Where am i gong wrong ?

REPLY [2 votes]: Every uniformly continuous function is bounded on every bounded interval included in its domain of definition. The proof only uses that the real line is Archimedean.