Defining a topology on a function space using a notion of convergence. Is it possible? Let $X$ be a space of real valued functions. I don't know if this $X$ is in general  a metric space.
I want to define a topology on $X$.
If I have only a notion of convergence on $X$, is there a way to define a topology on $X$ using this convergence?
Edit Does the possibility of defining a topology depend on the particular notion of convergence that I have on $X$?
Thanks!
 A: The answer is yes, though actually characterizing the topology in terms of open/closed sets may be complicated.
If the space $X$ is first-countable, then its topology is determined entirely by the convergent sequences in $X$ and their limits. This is because open-ness and closed-ness in first-countable spaces are properties of sets that can be phrased entirely in terms of convergence of sequences. In particular, if $X$ is first-countable, then a subset $C\subset X$ is closed if and only if for any sequence $(x_n)$ in $C$ satisfying $x_n\to x$, we have $x\in C$.
However, if $X$ is not first-countable then closed-ness and open-ness are no longer characterized by convergent sequences. However, these properties are still characterized by convergent nets. Nets are a generalization of the concept of a sequence. A set $C\subset X$ in an arbitrary topological space $X$ is closed if and only if for any net $(x_\alpha)$ whose elements are in $C$ with $x_\alpha\to x$, we have $x\in C$.
First countable spaces can essentially be thought of as topological spaces with enough regularity (in terms of countability axioms) to get by with sequences, instead of the full generality of nets.
So how can you define a topology on $X$ using convergent nets (or sequences if $X$ is first-countable)? First, come up with a criterion of convergence for nets. Nets can be defined independently of a topology: just like a sequence is just a function $\mathbb{N}\to X$, a net is just a function $A\to X$ where $A$ has a few order properties. Once you have characterized convergence for all nets, then take a subset $S\subset X$. Declare $S$ to be closed if and only if for any net $(x_\alpha)$ in $S$ that converges to $x$, it follows that $x\in S$. Use this to define all closed sets, and you have a topology. In a first-countable space, you can do everything described using sequences instead of nets.
This construction is something that is commonly done in analysis, especially with function spaces. For example, consider the space $C(E)$ of continuous real-valued functions on a subset $E\subset\mathbb{R}^n$. When $E$ is compact this space is often given a metric topology, with the metric given by the supremum norm, and convergence in this norm is equivalent to uniform convergence. However, if $E$ is not compact, then a continuous function on $E$ does not necessarily have a finite supremum (in absolute value), so the supremum norm is no longer well-defined. What is commonly done is to give $C(E)$ the topology of uniform convergence on compact sets: you declare a sequence of functions $(f_n)$ to converge to the limit $f$ if $(f_n)$ converges to $f$ uniformly on every compact subset of $E$.
The reason this method of defining a topology is useful is because sometimes, the most interesting question you can ask about a topology is precisely "which sequences/nets converge?" rather than "which sets are open/closed?" In the case of function spaces, it is more natural to ask if a net/sequence of functions converges to another function than to ask if a set of functions is open/closed, because typically the former conclusion is more directly useful.
Addendum: Also, continuous functions are characterized by their behavior on convergent nets/sequences, so understanding the notion of convergent nets/sequences allows you to understand the notion of continuous functions. This is often the more practical characterization of continuity: for example, it is usually much more efficient and common to verify that a given real-valued function on $\mathbb{R}^n$ is continuous by checking its behavior on sequences, as opposed to examining the preimage of an open set.
Addendum: I'm no set theorist or logician, so someone may have to correct me if I'm making a foundational error. I'm a bit iffy about the idea of "all nets." I'm pretty sure everything I've said is ok if $X$ is first-countable.
A: An important example in probability theory is almost sure convergence. That is, you have a probability measure $P$ on a $\sigma$-algebra $\cal A$ over your domain $\Omega$ and a sequence $f_n$ of measurable functions convergec almost surely to $f$ if $P(\lbrace \omega\in \Omega: f_n(\omega) \to f(\omega)\rbrace)=1$. Except for rather uninteresting cases there is no topology $\cal T$ on the space of measurable functions such that $\cal T$-convergence is a.s.-convergence.
The reason is that, in  every topological space, convergence $f_n\to f$ is equivalent to the fact that each subsequence has a further subsequence converging to the same limit.
This is not so for a.s. convergence because every sequence converging in measure (i.e. $P(\lbrace \omega\in \Omega: |f_n(\omega)-f(\omega)| > \varepsilon\rbrace \to 0$ for all $\varepsilon>0$) satisfies this property for a.s. convergence.
