Is $f(x) = 1/x$ over $[0.1 , 1]$ uniformly continuous but $1/x$ on $(0,1)$ is not? I know that there is a theorem that says that "Every continuous function on a bounded closed interval $[a,b]$ is uniformly continuous therein."
But I know that the function $f(x) = 1/x$ defined on (0,1) is not uniformly continuous (I know how to prove this). Does the previous theorem is saying that $f(x) = 1/x$ defined on [0.1,1] is uniformly continuous? why? what should I change in my proof of being not uniform continuous.
Here is my proof of not being uniformly continuous:
Let $\delta > 0,$ take $\epsilon = 1$ and $x = \min \{1, \delta\}$ and $a = \frac{x}{2}.$ then $|x - a | = |x - x/2| = \frac{x}{2} < \delta$ but $|\frac{1}{a} - \frac{1}{x}| = |\frac{2}{x} - \frac{1}{x}| = \frac{1}{x} > 1.$ And so $f$ is not uniformly continuous.
Also, my justification that it is continuous because it is the division of 2 polynomials (and in this justification I do not see any use of open or closed intervals)
Could anyone help me in answering my questions, please?
Thanks in advance!
 A: Uniform Continuous function
To understand it clearly, you should think what uniform continuity looks like. Let me illustrate it.
Given function is:
$f(x) = \frac{1}{x}$
Claim: In the interval (0, 1), you cannot find  $\delta$ corresponding to  any $\epsilon$.
To understand this, fix  $\epsilon = 1$, and choose whatever $\delta$ you think will satisfy the definition. You will find that this will not work. Reduce $\delta$ by half. You will find this will also not work. Keep on dividing it by half. You will  find  that no such $\delta$ works.
This happens because  $f$ grows too fast in the neighbourhood of 0 so that no such $\delta$ sized sub-interval of (0, 1) can capture the growth of $f$.
In the case of interval [0.1, 1], there is no such neighborhood where $f$ grows rapidly (which is true for continuous functions on closed and bounded intervals), therefore you can find a $\delta>0$ corresponding to every $\epsilon>0$.
How to find $\delta$ corresponding to $\epsilon$ in the interval [0.1, 1]:
Choose an arbitrary $\epsilon > 0$.
$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{|xy|}$
Using the fact that $x, y \geq 0.1$,
$|\frac{1}{x} - \frac{1}{y}| = \frac{|x-y|}{|xy|} \leq 100 |x-y| < \epsilon$
Clearly, $\delta  = \frac{\epsilon}{100}$
A: $$\left|\frac1{x-\delta}-\frac1x\right|=\left|\frac{\delta}{(x-\delta)x}\right|<\epsilon$$ requires
$$\delta<\frac{\epsilon x^2}{\epsilon x+1}<\epsilon x^2$$ so that no finite bound can hold for all $x$.
