In the following, I am trying to consider the (sequential) compactness methods in applications from a rather abstract point of view. I am not sure, whether such an abstraction is really meaningful.

Preliminary remark: Let $X$ be a set, $\tau_1$ and $\tau_2$ two topologies on $X$ and consider the join topology $\tau_1 \vee \tau_2$, i.e. the topology generated by $\tau_1 \cup \tau_2$. If $x \in X$ and $x_\alpha$ is a net in $X$ that $\tau_1$-converges to $x$ and $\tau_2$-converges to $x$ then $x_\alpha$ $(\tau_1 \vee \tau_2)$-converges to $x$.

I often meet compactness theorems in applications that establish $\tau$-convergence of some sequence $x_n$ (for some topology $\tau$) by convergence in some weaker $\tau_1$-topology together with some additional properties that need to be satisfied by the sequence $x_n$. Sometimes, these additional properties can be rephrased by relative (sequential) compactness in another weaker topology $\tau_2$.

As an example, consider $X := L^1(P)$ for some probability measure $P$, $\tau$ the topology of convergence in $L^1$-norm, $\tau_1$ the topology of convergence in measure and $\tau_2 := \sigma(L^1, L^\infty)$ the weak topology. The Vitali convergence theorem states that on $L^1(P)$ a sequence $x_n$ $\tau$-converges if and only if $x_n$ $\tau_1$-converges and $x_n$ is uniformly integrable. By Dunford-Pettis and Eberlein-Smulian, the uniform integrability of $x_n$ is nothing else but relative sequential compactness in the $\tau_2$-topology. In particular, the Vitali convergence theorem implies that $\tau = \tau_1 \vee \tau_2$.

Question 1: Does it also hold the other way round? I.e. if $x_n$ is $\tau_2$-convergent and $\tau_1$-relatively sequentially compact does it follow that $x_n$ is $\tau$-convergent?

Question 2: If Question 1 has a positive answer, does it always hold that if on some (possibly nice enough) topological space $(X, \tau)$ it holds that $\tau = \tau_1 \vee \tau_2$ then $\tau_1$-convergence of a sequence or a net and $\tau_2$-relative (sequential) compactness implies its $\tau$-convergence? (Note that in contrast to the above preliminary remark, we do not impose the $\tau_2$-convergence (it will then follow automatically)).

Question 3: It would be also interesting to know of such "additional properties" that can not be rephrased by a relative (sequential) compactness property.

  • $\begingroup$ What do you mean when you say a sequence is "relatively sequentially compact"? Do you just mean that every subsequence has a convergent subsubsequence? Also, what is the difference between Question 1 and Question 2? Is Question 1 supposed to be only about the example $X=L^1(P)$? $\endgroup$ – Eric Wofsey Jan 14 '16 at 1:01
  • $\begingroup$ Yes, rel. seq. compactness is as you described. Question 1 can be seen as a motivation for Question 2 and as a special case of Question 2 if Question 2 has a negative answer (as you have shown in your answer). $\endgroup$ – yadaddy Jan 14 '16 at 6:57
  • $\begingroup$ Oh, I didn't notice that you had swapped the roles of $\tau_1$ and $\tau_2$ in Question 1. $\endgroup$ – Eric Wofsey Jan 14 '16 at 6:59

Here is a counterexample to Question 2. Consider $X=[0,1)$ with $\tau$ its usual topology. Let $\tau_1$ be the coarser topology that glues $1$ to $0$ (i.e., neighborhoods of $0$ must contain intervals $(1-\epsilon,1)$, so $(X,\tau_1)$ is a circle) and let $\tau_2$ be the coarser topology that glues $1$ to $1/2$ (i.e., neighborhoods of $1/2$ must contain intervals $(1-\epsilon,1)$, so $(X,\tau_2)$ is like the letter P). It is easy to verify that $\tau=\tau_1\vee\tau_2$. But the sequence $(1-1/n)$ converges with respect to $\tau_1$ and $\tau_2$ (to $0$ and $1/2$, respectively), but does not converge with respect to $\tau$.

The additional hypothesis you need for this to work is that if a sequence $(x_n)$ converges $x$ with respect to $\tau_1$ and converges to $y$ with respect to $\tau_2$, then we must have $x=y$ (note that this hypothesis is certainly necessary if the topologies are Hausdorff, since such a sequence with $x\neq y$ will then give a counterexample as above). With that assumption, suppose $(x_n)$ $\tau_1$-converges to $x$ and is $\tau_2$-relatively sequentially compact. Then every $\tau_2$-convergent subsequence of $(x_n)$ must converge to $x$. So by relative sequential compactness, every subsequence contains a subsubsequence which converges to $x$ with respect to both $\tau_1$ and $\tau_2$ (and hence, with respect to $\tau$). It follows that the sequence $(x_n)$ converges to $x$ with respect to $\tau$.

In particular, this gives a positive answer to Question 1, and in fact shows that if $\tau_1$ and $\tau_2$ are Hausdorff, Question 2 has a positive answer for $(\tau_1,\tau_2)$ iff it has a positive answer for $(\tau_2,\tau_1)$.

(The result of the second paragraph also works with "sequence" replaced by "net" and "relatively sequentially compact" replaced by "relatively compact" everywhere.)

  • $\begingroup$ Very nice counterexample to Question 2 which precisely shows the necessity of your additional hypothesis (A) on $x_n$. Since (A) is symmetric in $\tau_1$ and $\tau_2$ your answer in the third paragraph is clear (Question 2 has a partial positive answer). So, the fact that (A) holds for $X = L^1(P)$ is basically due to the uniform boundedness principle: $\tau_2$-rel. seq. compactness implies $\tau$-rel. seq. compactness ($\tau = \tau_1 \vee \tau_2$ the norm topology) and with $\tau_1$-convergence to $x$ we have immediately $\tau$-convergence to $x$ (and thus (A) holds). $\endgroup$ – yadaddy Jan 14 '16 at 7:25
  • $\begingroup$ (I leave your answer unaccepted for some time since Question 3 is open.) $\endgroup$ – yadaddy Jan 14 '16 at 7:28

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