How to distinguish between "for all $x>0$", "on the interval $(0,\infty)$" and "on the interval $[a,\infty)$ for all $a>0$"? Assume in the following that $x,a$ are real numbers. How do I distinguish between the statements "for all $x>0$", "on the interval $(0,\infty)$" and "on the interval $[a,\infty)$ for all $a>0$"? As far as I can tell, these three sets are the same set, but in my assignment, there seems to be important differences. 
For instance in one exercise I want to prove that the series $S(x)$ "converges uniformly on the interval $[a,\infty)$ for all $a>0$". Later I am asked to prove that the same series does not converge uniformly "on $(0,\infty)$". Then later on, the lecturer claims that by picking an arbitrary closed interval $[a,b] \subset (0,\infty)$, and proving some statement about some series on that closed interval, concludes that the statement is true for all $x>0$ since $[a,b]$ is arbitrary.
What is going on? When I am faced with these statements, how should I interpret them? How do I distinguish between these statements and similar statements like them? Which of these statements imply the others? It would be best if you could generalize away from examples involving series.
 A: The key here is to observe that although
$$
\bigcup_{a>0} [a,\infty) = (0,\infty)
$$
the two statements


*

*$(S_n)_n$ converges uniformly on $(0,\infty)$

*$(S_n)_n$ converges uniformly on $[a,\infty)$
are not equivalent. Namely, the first implies the second, but there are cases (as, apparently, yours) where the second holds but not the first.
To see why, let us look at what the first statement means:

$(S_n)_n$ converges to (the function) $S$ uniformly on $(0,\infty)$: $$\sup_{(0,\infty)} \lvert S_n(x)-S(x)\rvert \xrightarrow[n\to\infty]{} 0$$

while the second statement says

Fix any $a>0$. Then $(S_n)_n$ converges to (the function) $S$ uniformly on $[a,\infty)$: $$\sup_{[a,\infty)} \lvert S_n(x)-S(x)\rvert \xrightarrow[n\to\infty]{} 0$$

Can you see the difference? It is the same sort of difference as in saying "for any fixed $a>0$, $\ln$ is bounded on the interval $[a,1]$" vs. "$\ln$ is not bounded on $(0,1]$".

Now, I assume the last part is something like showing that $S$ is continuous on $(0,\infty)$, "since all $S_n$'s are and you have uniform convergence on every $[a,\infty)$." The subtlety here is that while you don't have uniform convergence on $(0,\infty)$, the weaker statement is enough because continuity is a local property. Namely, you only need uniform convergence on some interval $(x_0-\varepsilon, x_0+\varepsilon)$ around $x_0>0$ to show continuity at $x_0$. And you have that by taking $a=\frac{x_0}{2}$.
A: The statement
$$\forall a > 0\colon f_n \text{ converges uniformly to } f \text{ on } [a,\infty)$$
is different than the statement
$$f_n \text{ converges uniformly to } f \text{ on } (0,\infty)$$
For example, let $f_n(x) = 1/(nx)$. Let $a>0$. Then $f_n(x)\leq 1/(na)$ for all $x\geq 0$. Therefore, letting $\epsilon > 0$ be arbitrary, we can pick $n$ large enough to ensure $\sup_{x\in[a,\infty)}|f_n(x)|<\epsilon$. However, this is not true on $(0,\infty)$.

What is concerning, however, is that it seems like the lecturer is
  trying to get you to show that the sequence is not convergent
  pointwise on $(0,\infty)$. This strikes me as odd since if a sequence
  converges uniformly on each set of the form $[a,\infty)$, it
  convergence pointwise on $(0,\infty)$. Perhaps there is a typo there.

