Convergence of Functions (indexed by an uncountable set) + application to a specific example, the convergence of indicator functions My specific question is, I believe the following two statements are correct. If they are, could someone please show me how to I prove the following rigorously (preferably using a $\epsilon, \delta$ proof). 
For each $t\in \mathbb{R}^+$ we have the following limits:
\begin{equation}
\lim_{s\downarrow\downarrow t}\mathbb{1}_{[0,s]} = \mathbb{1}_{[0,t]}
\end{equation}
and 
\begin{equation}
\lim_{s\uparrow\uparrow t} \mathbb{1}_{[0,s]} = \mathbb{1}_{[0,t)}
\end{equation}
where $\downarrow\downarrow$ and $\uparrow\uparrow$ are the right and left limits and $\mathbb{1}$ is the indicator function.
I intuitively feel as though these two statements are true because the arbitrary intersection of closed sets is necessarily closed, while the arbitrary union of closed sets is not necessarily closed (and in this case if we were to take $\bigcup_{n\in\mathbb{N}} [0,t-\tfrac{1}{n}] = [0,t)$). 
More generally - I believe that my confusion in proving this statement arises from the fact that if we were instead to deal with a sequence of functions, then I would know how to do this proof. E.g. I would be able to easily show that
\begin{equation}
\lim_{n\rightarrow\infty}\mathbb{1}_{[0,t+\frac{1}{n}]} = \mathbb{1}_{[0,t]}
\end{equation}
However as soon as I look at this statement, I realise that I am dealing with not just the convergence of a COUNTABLE sequence of functions. Rather I am dealing with the convergence of uncountable set of functions namely indexed by the real numbers.
So in saying that, perhaps a useful answer would be able to help me understand what it even means for this convergence/limit to hold in an $\epsilon,\delta$ sense. E.g. given a family of functions $\left\{f_t:\mathbb{R}\longrightarrow \mathbb{R}\mbox{ } \middle\vert\mbox{ } t\in \mathbb{R}^+\right\}$,what would it mean/what would be a rigorous definition for the statement:
\begin{equation}
\lim_{t\rightarrow s} f_t = f^*
\end{equation}
for some function $f^*:\mathbb{R}\longrightarrow \mathbb{R}$.
Thanks in advanced.
 A: It's all about the topology on the function spaces. By the way, the following are several options, of course there are much more.
1) Pointwise. Fix an $x\in{\bf{R}}$, given $\epsilon>0$, there exists some $\delta>0$ such that for all $t>0$ with $0<|t-s|<\delta$, then $|f_{t}(x)-f^{\ast}(x)|<\epsilon$.
2) Uniform. Given $\epsilon>0$, there exists some $\delta>0$, for all $t>0$ with $0<|t-s|<\delta$ and all $x\in{\bf{R}}$, then $|f_{t}(x)-f^{\ast}(x)|<\epsilon$.
3) Uniform on compact sets. Given $\epsilon>0$ and a compact set $K$ of ${\bf{R}}$, there exists some $\delta>0$, for all $t>0$ with $0<|t-s|<\delta$ and all $x\in K$, then $|f_{t}(x)-f^{\ast}(x)|<\epsilon$.
A: I'm assuming you're talking about pointwise convergence, since otherwise your limits don't exist.
$\displaystyle\lim_{s\downarrow\downarrow t}\mathbb{1}_{[0,s]} = \mathbb{1}_{[0,t]}$
 simply means that for any decreasing sequence $(s_n)_{n=1}^\infty$ which converges to $t$ and has at most finitely many occurrences of $t$ we have:
$$\lim_{n\to\infty} \mathbb{1}_{[0,s_n]} = \mathbb{1}_{[0,t]}$$
or more explicitly, that for every $x \in \mathbb{R}$ holds $\displaystyle \lim_{n\to\infty} \mathbb{1}_{[0,s_n]}(x) = \mathbb{1}_{[0,t]}(x)$.
Now this should be easy to prove. Indeed, fix a decreasing sequence $(s_n)_{n=1}^\infty$ with $\displaystyle\lim_{n\to\infty} s_n = t$.
For $x \le t$ we have $\mathbb{1}_{[0,t]}(x) = 1$ and also $\mathbb{1}_{[0,s_n]}(x) = 1$ for every $n \in \mathbb{N}$, since $s_n \ge t$. Hence, $\displaystyle \lim_{n\to\infty} \mathbb{1}_{[0,s_n]}(x) = \mathbb{1}_{[0,t]}(x)$.
On the other hand, for $x > t$ we have $\mathbb{1}_{[0,t]}(x) = 0$.
By defintion of $\displaystyle\lim_{n\to\infty} s_n = t$, for $\varepsilon = x -  t > 0$ there exists $n_0 \in \mathbb{N}$ such that for $n \in \mathbb{N}$ holds:
$$n \ge n_0 \implies s_n - t < x - t$$
We have ommited the absolute values since $s_n \ge t$. This implies $s_n < x$ for every $n \ge n_0$, so $x \notin [0, s_n]$ or equivalently $\mathbb{1}_{[0,s_n]}(x) = 0$. Again we get $\displaystyle \lim_{n\to\infty} \mathbb{1}_{[0,s_n]}(x) = \mathbb{1}_{[0,t]}(x)$.
We conclude $\displaystyle \lim_{n\to\infty} \mathbb{1}_{[0,s_n]}(x) = \mathbb{1}_{[0,t]}(x)$ for all $x \in \mathbb{R}$, so $\displaystyle \lim_{n\to\infty} \mathbb{1}_{[0,s_n]} = \mathbb{1}_{[0,t]}$ pointwise on $\mathbb{R}$.
Since the sequence $(s_n)_{n=1}^\infty$ was arbitrary, we have $\displaystyle\lim_{s\downarrow\downarrow t}\mathbb{1}_{[0,s]} = \mathbb{1}_{[0,t]}
$.
Analogously, \begin{equation}
\lim_{s\uparrow\uparrow t} \mathbb{1}_{[0,s]} = \mathbb{1}_{[0,t)}
\end{equation}
means that for every increasing sequence $(s_n)_{n=1}^\infty$ with $s_n \xrightarrow{n\to\infty} t$ and at most finitely many occurrences of $t$ holds $\displaystyle\lim_{n\to\infty} \mathbb{1}_{[0,s_n]} = \mathbb{1}_{[0,t)}$ pointwise on $\mathbb{R}$. The proof is similar to the above.
