I'm trying to prove that:

Let $\Omega \subset \mathbb{C}$ be an open set, $K$ a bounded connected component of $\mathbb{C} \setminus \Omega$. If $\,\{p_n(z)\}$ is a sequence of polynomials that converges uniformly on every compact subset of $\Omega$, then it converges uniformly on $K$.

By the maximum modulus principle, it suffices to prove that there exists a compact set $C \subset \mathbb{C} \,$s.t.$\, K\subset C\,$and$\,\partial C\subset \Omega$.

I reduced this problem to the assertion below:

Suppose $X \subset \mathbb{C}$ is a closed set, not connected. Let $K,L\,$ be two distinct connected components of $X$. (One may assume that $K$ is bounded.) Assume that $K$ is bounded.

Then there exist two disjoint closed sets $Y, Z \subset X$ satisfying $$K \subset Y,\, L \subset Z, \,\text{and} \,\,X = Y \sqcup Z .$$

How can I show this? Thank you.

Edited: My final answer.

Lemma 1.

Let $X$ be a Hausdorff space and $K\subset X$ have a compact neighbourhood $T$. Then $K$ is a connected component of $X$ if and only if $K$ is a connected component of $T$.

Proof: See here. $\square$

Put $X = \mathbb{C} \setminus \Omega \cup \{\infty\} \subset S^2$. Then $X$ is compact and Hausdorff.

Take $R>0$ so large that $K\subset B(0,R)$. Applying the Lemma 1 to $T = X \cap \overline{B(0,R)}$ one can verify that $K$ is a connected component of $X$.

Let $L$ be the connected component of $X$ that contains $\infty$. Then clearly $K, L$ are distinct connected components of $X$. Regarding $K$ and $L$ as quasi-components of $X$, one can choose two disjoint closed sets $Y, Z \subset X$ satisfying $$K \subset Y,\, L \subset Z, \,\text{and} \,\,X = Y \sqcup Z .$$ Then both $Y$ and $Z$ are closed in $S^2$, and thus by normality there exists an open set $U$ s.t. $$Y\subset U \subset \overline{U} \subset S^2 \setminus Z.$$ Since $\infty \notin \overline{U}$, $\overline{U}$ is a compact subset of $\mathbb{C}$. Moreover $\partial U \cap X = \emptyset$, i.e., $\partial U \subset \Omega$.

Then by maximum modulus theorem, we can show that $\{ p_n (z)\}$ is uniformly Cauchy on $K$!!

  • $\begingroup$ See math.stackexchange.com/questions/1537607/… $\endgroup$ – Robert Z Feb 28 '18 at 7:22
  • $\begingroup$ What is $Y?$ $\,\,$ $\endgroup$ – zhw. Feb 28 '18 at 21:25
  • $\begingroup$ Edited, thanks. $\endgroup$ – Hiro Wat Mar 1 '18 at 5:46
  • $\begingroup$ What you are trying to prove contradicts the fact that K is connected. $\endgroup$ – Kavi Rama Murthy Mar 1 '18 at 5:55
  • 1
    $\begingroup$ @JohnBrevik The existence of such Y and Z was the problem. $\endgroup$ – Hiro Wat Mar 2 '18 at 9:49

The claim you are trying to prove is false. Below is an example in the case when both $K$ and $L$ are unbounded. [What I wrote initially was an attempt to simplify my example and I simplified it so much that $X$ became connected.]

The idea is that if $X$ is a metrizable topological space then the space $Q$ of connected components of $X$ equipped with the quotient topology, need not be Hausdorff. In particular, two elements $q_1, q_2\in Q$ may fail to be separated by a partition of $Q$ into two clopen subsets.

Now, the example. Let $K$ and $L$ be, respectively, the vertical lines $x=1, x=+1$ in the xy-plane. Let $G_n$ be a sequence of pairwise disjoint "U-shaped curves" in the strip $-1<x<1$, such that each $G_n$ is homeomorphic to ${\mathbb R}$ and equals the union of the two vertical rays $x=\pm(1 -\frac{1}{n}), y\ge -n$ and the horizontal interval $$ -1+\frac{1}{n}\le x\le 1 -\frac{1}{n}, y=-n. $$ (I might draw a picture if and when I have time.)

Let $$ X:= K\cup L \cup \bigcup_{n\in {\mathbb N}} G_n. $$ This space is clearly disconnected. Furthermore, $K, L, G_n (n\in {\mathbb N})$ are its connected components. On the other hand, suppose that $X=Y\sqcup Z$ is a partition of $X$ in two closed subsets containing $K$ and $L$ respectively. If $Y$ contains infinitely many of the curves $G_n$ then its closure contains both $K$ and $L$, which is impossible. Similarly, $Z$ cannot contain infinitely many of the curves $G_n$. Hence, $Y\cup Z$ cannot equal the entire $X$.

You can see that the space $Q$ of connected components of $X$ consists of two points $[K], [L]$ and a sequence $[G_n]$ which converges to both $[K]$ and $[L]$, making $Q$ non-Hausdorff.

Edit. Now, your modified question, has positive answer.

I will need the notion of a quasicomponent of a topological space. Recall that a subset of a topological space is called clopen if it is both closed and open. The quasicomponent $X_x$ of a point $x$ in a topological space $X$ is the intersection of all clopen subsets containing $x$. Equivalently, quasicomponents of $X$ are the equivalence classes of the equivalence relation $\sim$ on $X$, where $x\sim y$ if and only if there is no partition of $X$ in tow clopen subsets, one containing $x$ and the other containing $y$.

The component $C_x$ of $x$ in $X$ is always contained in the quasicomponent $X_x$, but, quasicomponents, are, in general, strictly larger than components. However, if $X$ is compact and Hausdorff, then quasicomponents in $X$ are equal to components, see for instance here.

Proposition. Suppose that $X$ is a compact Hausdorff topological space. Then for any two distinct quasicomponents $A, A'$ of $X$ there exist clopen subsets $U, U'\subset X$ such that $$ A\subset U, A'\subset U', X=U\sqcup U'. $$

Proof. By the definition of a quasicomponent, for any two points $a\in A, a'\in A'$ there exists a pair of disjoint clopen subsets $U, U'\subset X$ such that $a\in U, a'\in U'$ and $X=U\cup U'$. By connectedness of $A, A'$, it follows that $A\subset U, A'\subset U'$. Clearly, both $U, U'$ are clearly also closed. qed

Back to your original question, where $K$ is a compact connected component of a closed subset $X\subset R^n$ and $L$ is another component of $X$. If $X$ is compact, then simply use the above theorem. Suppose that $X$ is noncompact.

Lemma. $K$ is a quasicomponent of $X$.

Proof. Let $B$ be a sufficiently large closed round ball in $R^n$ whose interior contains $K$. Then $K$ is a component of $X_B:=B\cap X$, hence, a quasicomponent of $X_B$. Hence, $K$ is the intersection of all its clopen neighborhoods in $X_B$. By compactness of $K$, one of these clopen neighborhoods $U$ will be contained in the interior of $B$. Hence, $U$ is also clopen in $X$. Therefore, all clopen neighborhoods of $K$ in $X_B$ which are contained in $U$ are also clopen in $X$. Hence, $K$ equals the intersection of its clopen neighborhoods in $X$. qed

Lemma. Let $X$ be a closed subset of $R^n$, let $K, L$ be distinct connected components of $X$, where $K$ is compact. Then there exists a pair of clopen subsets $U, V\subset X$ such that
$$ K\subset U, L\subset V, X= U\sqcup V. $$

Proof. We just proved that $K$ is a quasicomponent of $X$. Since $L$ is connected and distinct from $K$, and $K$ is the intersection of its clopen neighborhoods, one of the clopen neighborhoods (say, $U$) of $K$ in $X$ will be disjoint from $L$. Thus, for $V:= X- U$, we obtain: $$ K\subset U, L\subset V, X= U\sqcup V $$ and $U, V$ are closed. qed

  • $\begingroup$ I think the $X$ in this example is connected. Indeed, unless I'm making a big oversight, if $X$ is disconnected we can write $X$ as a disjoint union $Y\cup Z$, $Y$ and $Z$ both open and closed in the relative topology. Since $X$ is assumed to be closed, $Y$ and $Z$ are already closed in the ambient space. Just be sure that $Y$ contains $K$, and $Z$ contains $L$ when you choose them. $\endgroup$ – John Brevik Mar 1 '18 at 14:37
  • $\begingroup$ @JohnBrevik: See the update. $\endgroup$ – Moishe Kohan Mar 1 '18 at 16:56
  • $\begingroup$ A nice counterexample, thank you. I edited. However is there any proof or disproof when K is bounded? My problem is the case in which K is bounded. $\endgroup$ – Hiro Wat Mar 1 '18 at 17:46
  • $\begingroup$ I think such an obstacle arises from the point at infinity; the closure of your example in the Riemann sphere has the only one connected component! I should replace $\mathbb{C}$ by $S^2$, perhaps. $\endgroup$ – Hiro Wat Mar 1 '18 at 18:23
  • $\begingroup$ @HiroWat: OK, I finally worked out a proof of the claim that you needed in the case when $K$ is bounded. $\endgroup$ – Moishe Kohan Mar 2 '18 at 1:02

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