Difference in interval notation What is the difference between these two exercise questions (regarding the intervals)?

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*Show that for each $\epsilon > 0$, $f_n$ is uniformly convergent on $[\epsilon, \infty)$


*Show that $f_n$ is not uniformly convergent on $(0, \infty)$
I mean in the first interval we don't have the 0 because $\epsilon > 0$ and in the second we don't have the 0 because it is an open interval.
So what is the difference?
 A: [I gave a glib answer above as a comment, but hopefully this answer will give you some actual insight. :-} ]
For a cartoon version that might make the difference easier to see, replace "$f_n$ is uniformly convergent on" with "we can find a positive number that is strictly less than all numbers in".
What happens here, and probably what happens in your exercise, is that the truth of the statement requires you to find some number such that .... something is true about it. In a question like yours, about a sequence converging, I'll bet that you have to find some $N$ such that for all $i \gt N$ .... something holds.
So for each $\epsilon \gt 0$ you can find such an $N$ - and from now on I'm going to write it as $N_{\epsilon}$ to emphasize that it depends on $\epsilon$. But as your $\epsilon$ gets smaller, $N_{\epsilon}$ gets bigger and grows beyond any upper bound. So you can't find a single $N$ that will work for all $\epsilon$ values $\gt 0$. Which is what happens in your original question.
A: Consider the sequence functions $\{f_n\}$ defined on $\mathbb{R^+}$ by
$$f_n(x) = \dfrac{1}{nx}$$
Now for a fixed $\epsilon > 0$, $\{f_n\}$ converges uniformly on the interval $[\epsilon, \infty)$ because $|f_n(x) - f_m(x)| =  \frac{1}x|\frac{1}n - \frac{1}{m}| \leq $ $\frac{1}\epsilon |\frac{1}n - \frac{1}{m}|$ gets arbitrarily small for sufficiently large $n,m$ (i.e., $\{f_n\}$ is a Cauchy sequence). But if you consider the interval $(0, \infty)$, where $1/x$ gets arbitrarily large, the choice of $n$ and $m$ depends on the value of $x$.
