3
$\begingroup$

Building upon this question in Math.SE, I think the following might be rather of interest for MO.

In the literature on measure theory, probability and functional analysis the definition of a subset $A \subseteq X$ of a topological space $X$ to be relatively sequentially compact is not unique:

  1. $A$ is relatively sequentially compact in $X$ if its closure $\overline{A}$ in $X$ is sequentially compact, i.e. every sequence in $\overline{A}$ has a convergent subsequence (with limit in $\overline{A}$)
  2. $A$ is relatively sequentially compact in $X$ if every sequence in $A$ has a convergent subsequence with limit in $\overline{A}$.

In definition 2 it is not really necessary to explicitely demand the limit to be contained in $\overline{A}$ (the limit of any sequence in $A$ is always contained in $\overline{A}$ and even in the sequential closure of $A$).

Clearly, 1 $\Rightarrow$ 2 but for general topological spaces $X$ the converse does not hold as shown in the linked Math.SE thread. Another example can be also found in Meggonsin, "An Introduction to Banach Space Theory", p. 161, Exc. 2.15.

Remark: It seems to be not only the case that a few authors prefer some definition to the other but rather that roughly speaking half of the literature (I was reading so far) uses definition 1 and the other half definition 2. Both have advantages and disadvantages.

Question: Can the spaces $X$ in which 1 $\Leftrightarrow$ 2 for any subset $A \subseteq X$ be characterized by already familiar topological properties?

Note that for the weak topology on a Banach space $X$ these two definitions seem to be equivalent since I found versions of the Eberlein-Smulian theorem which equate weak relative compactness to weak relative sequential compactness with both definitions. But which property of the weak topology on $X$ is really used to equate these two definitions?

$\endgroup$
  • $\begingroup$ The standard sources that I have seen all use definition 2. As pointed out in the linked math.so item, with definition 1 you could have a sequentially compact set that is not relatively so. And in applying the Eberlein-Smulian theorem, there may be situations where you want to assume only the property in definition 2. $\endgroup$ – PassingThru Jan 8 '16 at 16:32
  • $\begingroup$ In Functional Analysis there is a widely used notion of an angelic space, introduced by Fremlin: this is a space in which a relatively countably compact space is relatively compact and compact subsets are Frechet-Urysohn. In such spaces both notions of sequential compactness seem to coincide. More information on angelic spaces can be found in the book [J.Kąkol, W.Kubiś, M.López-Pellicer, Descriptive topology in selected topics of functional analysis. Developments in Mathematics, 24. Springer, New York, 2011]. $\endgroup$ – Taras Banakh Aug 12 '16 at 19:40
1
$\begingroup$

One important class of spaces for which the two definitions mentioned in the post are equivalent, are the first-countable spaces. One of the most important properties of any first-countable space $(X,\tau)$ is that given a subset $A\subseteq X$, a point $x$ lies in the closure of $A$ if and only if there exists a sequence $(a_n)_{n\in\mathbb{N}}$ in A which converges to $x$. Using this characterization, it is not hard to prove that definition (2) implies definition (1).

$\endgroup$
  • $\begingroup$ Yes, this property is also satisfied for Fréchet-Urysohn spaces (a space $X$ is Fréchet-Urysohn if the sequential closure of any subset $A \subseteq X$ coincides with its closure $\overline{A}$). But does it follow from $1 \Leftrightarrow 2$ that $X$ is Fréchet-Urysohn? Recall that, an infinite-dimensional Banach space with the weak topology is not first-countable and I am not sure whether it is Fréchet-Urysohn but the fact $1 \Leftrightarrow 2$ seems to be still satisfied there. $\endgroup$ – yadaddy Jan 8 '16 at 10:06
  • $\begingroup$ Could you add a proof that for FU spaces, (2) implies (1)? $\endgroup$ – Boaz Tsaban May 7 '16 at 21:03

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.