# Definition for “relatively sequentially compact”

A topological space $X$ is sequentially compact if every sequence in $X$ has a convergent subsequence.

Let $X$ be a topological space and $A \subseteq X$. I've seen two definitions for $A$ to be relatively sequentially compact in $X$:

1. the closure $\overline{A}$ of $A$ in $X$ is sequentially compact, which means that every sequence in $\overline{A}$ has a convergent subsequence (with limit in $\overline{A}$).
2. every sequence in $A$ has a convergent subsequence with limit in $\overline{A}$.

Clearly, 1 => 2.

Are these definitions equivalent? (I don't see how to reduce the sequence in $\overline{A}$ to a sequence in $A$, so it seems that they are not equivalent unless $X$ is something like a Fréchet-Urysohn in which points in the closure $\overline{A}$ can be approximated by a sequence in $A$. Then we could try to perform a diagonal argument.) If they are not equivalent, what is the right definition for abstract topological spaces $X$?

There may be contexts where the first definition is appropriate, but it does seem somewhat pathological in that a sequentially compact subspace may not be relatively sequentially compact. In this respect it differs from the second definition.

For example, take the Tychonoff plank $X = ([0, \omega_1] \times [0, \omega]) \setminus (\omega_1, \omega)$ and the subspace $A = [0, \omega_1) \times [0, \omega]$. Then $A$ is sequentially compact, since any sequence is confined to a subspace $[0, \alpha] \times [0, \omega]$ with $\alpha < \omega_1$, which is compact and first countable. On the other hand $\overline{A} = X$, which is not sequentially compact since $\{ (\omega_1, n) \}_{n=0}^\infty$ has no cluster point.

To decide which is the most useful definition would probably involve looking at a large number of applications, which I don't have available. I could even imagine some use for a definition $1\frac12$: a subspace $A$ of a topological space $X$ is relatively sequentially compact if there is a sequentially compact $B \subset X$ such that $A \subset B$. This is strictly weaker than definition 1 and stronger than definition 2, although I don't know if it is strictly stronger than definition 2.

• Thanks for the example. Up to now, I only met the necessity to introduce the notion of a "relatively sequentially compact" set in the context of compactness methods in functional analysis and probability theory for convergence of sequences in particular function spaces, e.g. Eberlein-Smulian or Prokhorov theorem. I saw say half of the authors using definition 1 and the other half using definition 2. As an example, I saw Eberlein-Smulian stating that weak compactness in Banach spaces is equivalent to rel. seq. comp. as in definition 1 in some literature and to definition 2 in other literature. – yadaddy Dec 23 '15 at 15:17
• I could find no actual use of either definition in a context where the other one would not work, so I would guess that you are free to pick the most convenient one. (probably number 2) – Niels J. Diepeveen Dec 24 '15 at 13:52

Well, in a general topological space there are different ways to define convergent series. For instance you could require the limit to be unique or not.

I think series in a general topological space are not as usefull as, say, in metric spaces. In topologcial spaces you get the "same" criterions for continuity, closedness, compactness involving series, if you use nets or filters instead of series.

So I do not know if there is a right definition. But I guess it depends on what you want. Probably, there is not the right choice.

And no your definitions are not equivalent. Just take the intervall $[-1,1]$ with infinitely many origins as $X$. That is $[-1,1] \times \mathbb{N}$ modulo the equivalence relation $(x,n)\sim (x,n')$ for $x\ne 0$. And choose $A=[-1,1]\times \{0\}$. Then $\bar{A}=X$

• Thanks for the example. These two definitions are mostly used in the context of functional analysis. So, let us assume at least some separation like Hausdorfness and complete regularity for $X$. – yadaddy Dec 22 '15 at 16:27
• But wait, sequences in $X$ do have convergent subsequences, since they do in $[-1,1]$ and all points in the origins are identified, e.g. $(x,n)$ converges to any point $(x,k)$ an $n \to \infty$. – yadaddy Dec 22 '15 at 17:13
• Sorry I assumed that the limit has to be unique which is not common. – user60589 Dec 22 '15 at 17:22