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I've encountered the sentences "topologized by convergence in probability" or "topologized by uniform convergences" and I would like to get an idea of the big picture.

Given notion of convergence on some set $X$ for which no topology has yet been specified (hence one cannot define convergence in the usual way), how it's possible to define a topology on $X$?

In this answer is said that it's possible to axiomatize the notion of convergence even without nets (and I'd be happy to not involve nets). Suppose one has a similar notion, it's enough to declare the closed sets as the sequentially closed to get a topology?

Gabriele Cacchioni
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First, you need a topological version of convergence. I could impose that a sequence of real numbers $(x_n)$ converge to $0$ if $x_1$ is odd. This is not a topological notion of convergence because the constant sequence $(0,0,0,0,\ldots)$ does not converge to $0$. (And we know that, in topology, a constant sequence must converge to the constant.) As Lucio points out, there are incredibly useful forms of convergence that are not topological, such as almost sure (or almost everywhere) convergence of a sequence of random variables (or measure functions).

We will describe a topological notion of convergence in terms of nets for the sake of generality. (One could also use filters.) If we describe convergence in terms of sequences, we are restricting how many topologies we could create because there are topologies in which sequential convergence is not sufficient.

The following is from Problem 11D in Willard's General topology textbook.

Suppose we have some notion of convergence on a set $X$ satisfying the following properties. Fix $x\in X$ and let $I$ be a directed set.

(a) If $x_i=x$ for each $i\in I$, then the net $(x_i)$ converges to $x$.

(b) If $(x_i)$ converges to $x$, then every subnet of $(x_i)$ converges to $x$.

(c) If every subnet of $(x_i)$ has a subnet converging to $x$, then $(x_i)$ converges to $x$.

(d) (Diagonal principle) If $(x_i)$ converges to $x$ and, for each $i\in I$, a net $(x^i_j)_{j\in J_i}$ converges to $x_i$, then there is a diagonal net converging to $x$; i.e., the net $(x^i_j)_{i\in I,\,j\in J_i}$, ordered lexicographically by $I$, then by $J_i$, has a subnet which converges to $x$.

Then if the closure of a subset $E$ of $X$ is defined by $$ \overline{E}:=\{x\in X \mid x_i\to x\ \text{for some net $(x_i)$ contained in $E$}\}, $$ the result is a topological space in which the notion of net convergence is as originally specified.

The moral of the story is that a topological notion of convergence tells you which sets are closed, and thus which sets are open. So we have a topology. In analysis, it is very common to describe and use topologies in this way. Since working with the open sets directly may be difficult, you may recall some useful theorems that rely only on convergence.

If $X$ and $Y$ are topological spaces and $f:X\to Y$ is a function, then $f$ is continuous iff whenever a net $(x_i)$ in $X$ converges to a point $x$ in $X$, then $f(x_i)\to f(x)$ in $Y$.

If $X$ can be described by sequences (which is true of first countable or metrizable spaces), then the above holds with nets replaced by sequences.

A topological space $X$ is compact iff every net in $X$ has a subnet which converges.

If $X$ is metrizable, then the above holds with net and subnet replaced by sequence and subsequence.

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John Griffin
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  • Let me introduce "(e)":

    (e) if $x_i\to x$ and $x_i\to y$, then $x=y$.

    Should we add (e) to the list of axioms of a "convergence notion" in order to get that the convergence notion induced by the (topology induced by notion of convergence) coincides with the originally specified notion of convergence?

    Just with (a), ..., (d) I was able to prove that if a net converges in the original notion, then it converges in the induced topology. But I could not prove that the notion of the topology gives back the original notion without using (e) at some point...

     – Derso Apr 07 '18 at 19:09
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    @AndersonFelipeViveiros Such a property is not always true. In fact, property (e) is equivalent to the topology being Hausdorff. Thus, by imposing such a condition, you are restricting the obtainable topologies to only the Hausdorff ones. – John Griffin Apr 07 '18 at 19:15
  • Pls, help me find the subnet in the diagonal principle.The set $U(x)$ of opens containing $x$ is directed with the preorder: $U\prec V$ if $V\subset U$. $\forall U\in U(x)$, $\exists i_U\in I$ s.t. $i_U\prec i\implies x_i\in U$. Then, $\exists j_U\in J_{i_U}$ s.t. $j_U\prec j\implies x^{i_U}_j\in U$. I wanted the map $h:U(x)\to \scr D$ ($\scr D$ the lexic. directed set) $h(U):=(i_U,j_U)$ were mon. cofinal. I need that $U\prec V\implies (i_U,j_U)\prec (i_V,j_V)$. Can $i_U$, $j_U$ be chosen for this? Furthermore, I need that $\forall (i,j)$ $\exists (i,j)\prec (i_U,j_U)$... – Derso Apr 07 '18 at 23:12
  • This would be possible if I could choose "minimals" $i_U$ and $j_U$... – Derso Apr 07 '18 at 23:31
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    @AndersonFelipeViveiros I don't think you will be able to proceed like that. An easy solution is to change your directed set. Consider the set of all triples $(U,i,j)$ such that $U$ is a neighborhood of $x$ and $x_i,x^i_j\in U$, and partially order this set by $(U,i,j)\le (V,i',j')$ if $V\subseteq U$ and $i\le i'$ and $j\le j'$. Then define $h(U,i,j)=(i,j)$. – John Griffin Apr 08 '18 at 00:24
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    Thanks for the idea! – Derso Apr 08 '18 at 00:47
  • How is this "If $X$ can be described by sequences (in particular, it is metrizable), then [...]"? Could you make this a bit more precise? (for example, if the sequence convergence notion has not the uniqueness of the limit property, then $X$ cannot be metrizable, since every metric space is Hausdorff). – Derso Apr 14 '18 at 06:38
  • https://math.stackexchange.com/questions/2736416/if-x-can-be-described-by-sequences-then-it-is-metrizable – Derso Apr 14 '18 at 07:16
  • @AndersonFelipeViveiros The sentence you quoted isn't meant to say that every space that can be described by sequences is metrizable - only that metrizable spaces can be described by sequences. Hence the words "in particular" and not something like "equivalently". – John Griffin Apr 14 '18 at 19:45
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These topologies are metrizable. That is, there is a metric such that every open set is a union of open balls obtained from this metric. In a metric space, a set $S$ is closed if and only if the limit of every convergent sequence of elements of $S$ is in $S$. So for metrizable topologies, itis enough to specify the convergent sequences.

This does not apply to all topological spaces. For example, if $X$ is an uncountable set, there is a nonmetrizable topology consisting of the empty set and subsets with an at most countable complement. This topology differs from the discrete topology, but the same sequences converge (those that are eventually constant).

In purely topological terms, a condition that ensures that a set is closed if and only if it is sequentially closed is that that every point has a countable neighborhood base.

Michael Greinecker
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  • "These topologies are metrizable. That is, there is a metric such that every open set is a union of open balls obtained from this metric". Do you mean that every topology obtained from a convergence notion is metrizable? – Gabriele Cacchioni Oct 28 '17 at 15:23
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    @GabrieleCacchioni No. But if you know a topology is metrizable, it is enough to know which sets are sequentially closed to recover the topology. – Michael Greinecker Oct 28 '17 at 15:35
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I don't have a complete answer for you, but these are some things that might help you:

1) There are notions of convergence which are not induced by a topology. One example is the almost sure convergence (at least in $\mathbb{R}^n$ with lebesgue measure).

2) Different topologies can lead to the same notion of convergence (I'm talking about convergence for $\textit{sequences}$). One example is the space $l^1$ with respectively weak and strong topologies. However, in this case only in one of the two topologies closed and sequentially closed sets coincide.

If you have a notion of convergence that satisfies reasonable properties (I think that it suffices to impose a) any constant sequence converges to that constant, b) if a sequence converges, than any subsequences converges to the same limit) then you can define closed sets to be those containing all their limits points with respect to this notion of convergence. This notion of closure satisfies the axioms ($\emptyset$ and the space are closed, finite unions and arbitrary intersection of closed are closed). Therefore this notion of convergence defines a topology in this way. The problem now is that the topology defined this way induces $\textit{another}$ notion of convergence. So the point is: do these two notions of convergence actually coincide? In order for them to coincide you probably need to impose more axioms on the notion of convergence. I'm sorry but I don't know exactly which these axioms are (probably a quite relevant one is: the sequence converges, say to $l$, if and only if any subsequence contains a subsequence converging to $l$; this is the property violated by the almost sure convergence). Note that if the two notions coincide, then you would have that closed sets and sequentially closed sets are the same, so not all topologies can be defined this way.

Lucio
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This is a special case of the construction of the final topology: a sequence $f : \mathbb N \to X$ converges to $x$ iff the map $$\begin{align*} \bar f : \mathbb N \cup \{\infty\} &\to X \\ n & \mapsto f(n) \\ \infty &\mapsto x \end{align*}$$ is continuous, where $\mathbb N \cup \{\infty\}$ is given the topology whose open sets either:

  • do not contain $\infty$
  • contain $\infty$ and have finite complement

The topology determined by a family of formally convergent sequences $f_i$ with formal limits $x_i$, is then simply the final topology for the $\bar f_i$.

Similarly, a net $f : N \to X$ converges to $x$ iff the map $$\begin{align*} \bar f : N \cup \{\infty\} &\to X \\ n & \mapsto f(n) \\ \infty &\mapsto x \end{align*}$$ is continuous, where $N \cup \{\infty\}$ is given the topology whose open sets either:

  • do not contain $\infty$
  • contain $\infty$ and contain an "arrow" of the form $\{n \geq n_0\}$ for some $n_0 \in N$.

Bonus: Thinking about sequences as continuous maps is helpful in other situations too, for example, when one wants to characterize when a sequence converges in, for example, a product space, or more generally an initial topology. The proof becomes a one-liner: by the universal property of the initial topology. Compare with Convergence in the product topology iff mappings converge

Bart Michels
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