I am reading the book "Z. Semadeni, Banach Spaces of Continuous Functions". At Definition 3.6.11, he defines a generic notion of convergence as follows:

enter image description here

By "upward filtering ordered sets" he means a "directed set", so his notion of net is the usual one.

Imediatelly I noticed that some requirements are missing in order to obtain some basic properties on convergent nets in topological spaces, such as:

(a) If a net is convergent then any of its subnet is convergent and they have the same set of limit points;

(b) If two convergent nets coincide in a cofinal subset of its index set, then they have at least one limit point in common;

among others. So, surely, I though: not every convergence of this kind is given by a topology. However, his comment on the almost everywhere convergence of the Lebesgue measure (highlighted in red) intrigued me, since this kind of convergence satisfies properties (a), (b) and some other properties of nets in vector topological spaces. So, here are my questions:

(1) Why is the almost everywhere convergence on the Lebesgue measure not given by any topology?

(2) Are there some sufficient conditions which guarantees that a kind of convergence is given by a topology?

  • 1
    $\begingroup$ Your first question is answered in this PDF of E.T. Ordman, ‘Convergence Almost Everywhere Is Not Topological’, The American Mathematical Monthly, Vol. 73, No. 2 (Feb., 1966), pp. 182-183. $\endgroup$ May 20, 2020 at 20:46
  • $\begingroup$ @BrianM.Scott, you are a topology life-saver! What about the second question? We must have that "a net $(x_i)_{i\in I}$ converges to $x$ if, and only if, any subnet of $(x_i)_{i\in I}$ admits a subnet convergent to $x$". Would that be sufficient? $\endgroup$ May 20, 2020 at 20:51
  • $\begingroup$ I really don’t know. The references suggested in answers to this question might be of use; a quick look at Henno’s first reference suggests that it might be a good starting point. $\endgroup$ May 20, 2020 at 21:07
  • $\begingroup$ @BrianM.Scott True, though it is mostly about filters, as is most of the literature on convergence spaces; nets get ugly rather quickly, IMHO, with the proliferation of index sets etc. $\endgroup$ May 20, 2020 at 22:23
  • 1
    $\begingroup$ Ah Semadeni.. Nice book. Long time no read (I should have bought it, perhaps). $\endgroup$ May 20, 2020 at 22:39

1 Answer 1


Kelley General Topology (p. 74, Thm 9) mentions the following 4 conditions on net convergence to define a "convergence class" (on a set $X$):

a. If $S$ is a net such that $S_n = s \in X$ for each $n$ ($n \in N$, some directed set), then $S$ converges to $s$.

b. If $S$ converges to $s \in X$, then so does every subnet of $S$ (subnet in Kelley's sense, of course, defined on p. 70 of the book).

c. If $S$ does not converge to $s \in X$, then there is a subnet of $S$ no subnet of which converges to $s$. (A condition that a.e. convergence fails to fulfill, as shown by this argument by Ordman).

d. (what he refers to as theorem 2.4 on iterated limits): Let $D$ be a directed set, let $E_m$ be a directed set for each $m \in D$, let $F= D \times \prod_{m \in D} E_m$ (in the product (i.e. coordinatewise) ordering, which is directed too) and for each $(m,f) \in F$ let $R(m,f)=(m,f(m))$. If $\lim_m \lim_n S(m,n)=s$ ($S$ is a function into $X$ defined on all pairs $(m,n)$ with $m \in D \land n \in E_m$) for some $s \in X$ (according to the "convergence rule"), then $S \circ R$ converges to $s$ too.

The last condition is a technical one to ensure that the set of limits of nets from a set $A$, will be a valid idempotent ($\overline{\overline{A}} = \overline{A}$) closure operator. Read the book for details. All conditions a-d are valid in a topological space convergence, and if a convergence class obeys them we can define a topology for which this convergence is the topology-defined convergence.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .