1

Let $K\subset\overline{\Bbb C}=\Bbb C\cup\{\infty\}$ be a continua, that is non empty, connected and compact subset. I am reading a paper in which it is stated that there exists a sequence $\{a_n\}_{n\ge1}\subset\Bbb C$ such that $K$ is its cluster set. Thus, I think, fixed an arbitrary point $z_0\in K\cap\Bbb C$, we should find a subsequence $\{n_k\}_k$ such that $a_{n_k}\to z_0$.

But, why is this true?

Joe
  • 11,745
  • 1
    Make a grid of size $1/2$ and take the vertices of the squares that intersect $K$. Then a grid of size $1/3$ and take the new vertices (if there are any) from squares that intersect $K$, and so on ... –  Apr 25 '18 at 13:51
  • Ah, in case the ambient space is source of concern. If $\infty\in K$, just move $K$ to the finite points by applying $z\mapsto 1/(z-a)$, for $a\notin K$, before constructing the sequence coming from those grids, and then move the sequence back applying $z\mapsto 1/z + a$. –  Apr 25 '18 at 14:00
  • 1
    Essentially the same ideas used Given closed $C \subseteq \mathbb{R}$ find a sequence with subsequences convergent to every point in $C$ and nowhere else will work for the more general case when only "nonempty and compact" are assumed, which incidentally is the most general possible case, since for each sequence the set of its subsequence limits is nonempty and compact. I'm a bit too busy now to write up a full answer, in case someone else was hesitating on my behalf. – Dave L. Renfro Apr 25 '18 at 16:06
  • Notice also that all constructions in the page that Renfro linked fail if you require the points of you sequence to be different (which is a strengthening of this problem requiring a countable set rather than a sequence that accumulates at $K$). The reason is that in that problem they want a sequence of elements of $C$. Makholm's construction can be salvaged by taking the end-points of the intervals, which translates to the vertices of the squares in the grids I mentioned, instead of taking points in $I\cap C$. –  Apr 25 '18 at 16:54

1 Answers1

1

The space $X=\overline {\Bbb C}$ is metrizable and separable because it is homeomorphic to $S^2.$ Any subspace of a separable metrizable space is separable. So there exists a countable $S\subset K$ such that $Cl_K(S)=K.$

Now $K$ is compact and $X$ is Hausdorff, so $K$ is closed in $X.$ So for any $T\subset K$ we have $Cl_K(T)=Cl_X(T).$ In particular, $Cl_X(S)=Cl_K(S)=K.$

If $K$ has more than $1$ member then any $x\in K$ belongs to $Cl_X(S\setminus \{x\})=Cl_K(S\setminus \{x\}),$ otherwise $K=Cl_K(S)=Cl_K(\{x\})\cup Cl_K(S\setminus \{x\})=\{x\}\cup Cl_K(S\setminus \{x\})$ is the union of two disjoint non-empty sets that are closed in $K,$ making $K$ disconnected.

So if $K$ has more than $1$ point then for every $x\in K$ there is a sequence of members of $S\setminus \{x\}$ that converges to $x.$

DanielWainfleet
  • 57,985