Let E be a finite dimensional Euclidean space. A well known theorem of Sierpinski states that if C is an infinite compact-i.e. closed and bounded-connected subset of E, then C cannot be the countable union of pairwise disjoint closed subsets of E. I have also seen a theorem of Hausdorff which implies that the same conclusion continues to hold if C is any infinite closed connected subset of E-not necessarily compact-provided C is locally connected. My question is: Does Hausdorff's theorem continue to hold, if one drops the requirement that C should be locally connected?..........Some of the statements I have read in the literature seem to suggest that the answer is "Yes" whenever the infinite connected subset C of E is locally compact. This will certainly be the case if E is a finite dimensional Euclidean space and C is a closed subset of E. But I am not really clear about the whole situation.
$\begingroup$
$\endgroup$
1
asked Mar 17, 2017 at 19:20
-
1$\begingroup$ I don't have time now to look into this, but I suggest googling Sierpinski "Un théorème sur les continus". For instance, I found this and this right away. $\endgroup$– Dave L. RenfroCommented Mar 17, 2017 at 19:43
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Sierpinsky himself constructed an example of a closed connected subset of $R^3$ which is a countable union of pairwise disjoint closed nonempty subsets.
W. Sierpinski, Sur guelgues proprietes topologiques du plan, Fund. Math. 4 (1923), 1-6.
answered Mar 18, 2017 at 12:59
-
$\begingroup$ This is the answer I was looking for. It shows that additional conditions on C besides being closed in E are needed, in order for Sierpinski's conclusion to follow. $\endgroup$ Commented Mar 19, 2017 at 19:47