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In a paper I am reading (not a published one), the following is considered obvious:

Let $K$ be a compact and connected subset of $\,\mathbb R^2$, with $\mathbb R^2\smallsetminus K$ connected, and $U\subset \mathbb R^2$ open with $K\subset U$. Then there exists a simply connected and open $V\subset \mathbb R^2$, with $K\subset V\subset U$. More generally, if $K$ is compact, $\mathbb R^2\smallsetminus K$ is connected and $U\subset \mathbb R^2$ open with $K\subset U$, then there exists an open $V\subset \mathbb R^2$, with $K\subset V\subset U$, such that all the connected components of $V$ are simply connected.

I have not managed to see why this is obvious. So far, I have shown this for simply connected compact sets $K$ with sufficiently smooth boundaries. Any ideas?

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  • $\begingroup$ If $K$ were not simply connected, wouldn't it have a hole, making $\mathbb R^2\setminus K$ not connected? $\endgroup$ – Nishant Feb 14 '15 at 18:49
  • $\begingroup$ @Nishant: Correct point. I added this piece of information to avoid misunderstandings, and in particular, path-connectedness. $\endgroup$ – Yiorgos S. Smyrlis Feb 14 '15 at 18:53
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This is related to "Can the complement of a simply connected set in $\bar{\mathbb{C}}$ in an open set always be covered by a simply connected union of balls?" In fact, Trevors proof needs to be adapted only slightly.

Since $K$ is compact, we can choose $\epsilon > 0$ and a finite set $\{x_1, \ldots, x_n\} \subseteq K$ such that $K$ is contained in the open set $$ V_0 := \bigcup_{i=1}^n B_\epsilon(x_i) $$ and $V_0$ is contained in $U$.

Using connectedness of $K$, it can be shown that $V_0$ is path-connected, but $V_0$ will not always be simply connected. However, we can construct a simply connected open subset $V$ of $V_0$ that contains $K$ in the following way.

Note that $V_0$ is bounded. Therefore its complement $\mathbb R^2 \setminus V_0$ has only one unbounded component $E$. By induction on $n$, it should be easy to prove that $\mathbb R^2 \setminus V_0$ has only finitely many bounded components $I_1, \ldots, I_k$. By assumption, $\mathbb R^2 \setminus K$ is open and connected. Thus it is path-connected. Since $E$ and $I_i$ are subsets of $\mathbb R^2 \setminus K$, it follows that for every $i \in \{1, \ldots, k\}$ there is a path $\gamma_i: [0,1] \to \mathbb R^2 \setminus K$ such that $\gamma(0) \in I_i$ and $\gamma(1) \in E$. Note that $$ \Gamma := \bigcup_{i=1}^k \gamma_i([0,1]) $$ is compact. I try to illustrate the construction above:

Note that the pictured example does not contain all possible pathologies.

Now, let $V$ be the unique connected component of $V_0 \setminus \Gamma$ that contains $K$. (The example above shows that $V_0 \setminus \Gamma$ does not need to be connected.)

We show that $V$ is open. Let $y$ be a point in $V$. Then $y$ is contained in $V_0 \setminus \Gamma$. Since $V_0$ is open and $\Gamma$ is compact, there is a $\delta > 0$ such that $B_\delta(y) \subseteq V_0 \setminus \Gamma$. Since $B_\delta(y)$ is connected, it follows that $B_\delta(y) \subseteq V$. Therefore $V$ is open.

It is easy to see that the unbounded component $E'$ of $\mathbb R^2 \setminus V$ contains the unbounded component $E$ of $\mathbb R^2 \setminus V_0$. It follows that $\Gamma \subseteq E'$ and $I_i \subseteq E'$ for all $i \in \{1, \ldots, k\}$. Thus $\mathbb R^2 \setminus (V_0 \setminus \Gamma) \subseteq E'$. It follows that every bounded component $I'$ of $\mathbb R^2\setminus V$ is a component of $V_0 \setminus \Gamma$. Thus by connectedness of $V_0$, it touches $\Gamma$, which implies $I' = E'$ contradicting the boundedness of $I'$. Therefore $\mathbb R^2 \setminus V$ is connected. This suffices to prove the Riemann mapping theorem for $V$. The following link explains how this is done in Ahlfors' Complex Analysis: http://math.ucr.edu/~res/math205B/ahlfors.pdf. So $V$ is biholomorphic to the open unit disk and we conclude that $V$ is simply connected. This ends the proof.

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  • $\begingroup$ If I understand correctly, the idea here (modulo technicalities) is: cover $K$ by balls contained in $U$, then identify the holes in the resulting set $V_0$ using its fundamental group, and cover them. $\endgroup$ – Daniel Robert-Nicoud Jan 9 '16 at 16:46
  • $\begingroup$ @Daniel Actually, no algebraic topology is needed. The idea is to connect all the components of $\mathbb R^2 \setminus V_0$ through paths and choose a connected open set $V$ that is not intersected by those paths such that $K \subseteq V \subseteq V_0$ and $\mathbb R^2 \setminus V$ is connected. The proof of simple connectedness of $V$ uses Complex Analysis. $\endgroup$ – user296355 Jan 9 '16 at 16:55
  • $\begingroup$ @DanielRobert-Nicoud In general, it is not possible to just cover the holes of $V_0$, because they may not be contained in $U$. That's why the proof has to use a different idea. $\endgroup$ – user296355 Jan 9 '16 at 17:04
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I want to go at it with complex analysis. Probably overkill, but potentially quite short. I'll use two facts:

  • Any open subset $X \subset \overline{\mathbb{C}}$ is simply connected if and only if both $X$ and its complement are connected.

  • Let $X \subset \overline{\mathbb{C}}$ be open and simply connected in $\overline{\mathbb{C}}$. If its complement has at least two points, then there is a biholomorphism $g : X \to B(0,1)$ (Riemann mapping theorem).

If $K$ is a point, then the result is obvious (take a small ball centered around $K$). Otherwise, we can see $K$ as a connected compact subset of the Riemann sphere $\overline{\mathbb{C}}$. Then $\overline{\mathbb C} \setminus K$ is a connected open subset of the Riemann sphere. By the first fact, $\overline{\mathbb C} \setminus K$ is simply connected. By the second fact, and since $K$ has at least two points, there is a biholomorphism $g : \overline{\mathbb C} \setminus K \mapsto B(0,1)$.

Since $\overline{\mathbb{C}} \setminus U$ is compact in $\overline{\mathbb C} \setminus K$, its image by $g$ is compact in $B(0,1)$. Hence, there exists $\varepsilon > 0$ such that $g(\overline{\mathbb{C}} \setminus U) \subset \overline{B}(0,1-\varepsilon)$.

Take $V := g^{-1} \bigl( \overline{B}(0,1-\varepsilon)^c\bigr) \cup K$, and delete the infinity if it happens to be in this set. Then $\overline{\mathbb C} \setminus V$ is biholomorphic (with $g$) to the disk $\overline B(0,1-\varepsilon)$, which ensures that is is connected (and stays connected if you have to delete the infinity).

The motivation of this proof is that, with the conformal mapping theorem, I can make the complementary of $K$ look like a disk, which allows me to shrink it a little without having to worry about potential pathologies.

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  • $\begingroup$ I very much like your approach using complex analysis. However, I wonder why $K$ is contained in the interior of the set $V$ that you constructed. This seems to be necessary to prove that $V$ is open and simply connected. Maybe I am missing something? $\endgroup$ – user296355 Jan 10 '16 at 16:59
  • $\begingroup$ @cbliz: the set $g^{-1} (\overline{B} (0,1-\varepsilon)^c))$ is (homeomorphic to) an annulus; its complementary has two connected components, and $K$ is one of them. Working on $B(0,1)$, you can easily see that any limit point of elements of $V^c$ is in $V^c$ or in $g^{-1} (S (0,1-\varepsilon)^c)$. $\endgroup$ – D. Thomine Jan 10 '16 at 17:16
  • $\begingroup$ @d-thomine Thanks for your reply. I see that $g^{-1}(\overline B(0,1-\varepsilon)^c)$ is homeomorphic to the annulus $\overline B(0,1-\varepsilon)^c$. But how does this imply that its complement has two components? An annulus in $\mathbb R^3$ does not have this property. I guess you need to extend $g^{-1}$ to a map that is surjective on $\overline{\mathbb{C}}$? $\endgroup$ – user296355 Jan 10 '16 at 18:32
  • $\begingroup$ @cblitz: There probably a theorem, bu here, you can see it by the construction. $\overline{C}$ is divided in three sets : $K$, $U_1 := g^{-1} (\overline{B} (0, 1-\varepsilon)^c)$, and $U_2 := g^{-1} (\overline{B} (0, 1-\varepsilon))$. The three are connected, be it by hypothesis or by construction. $\endgroup$ – D. Thomine Jan 10 '16 at 19:17
  • $\begingroup$ @d-thomine That is convincing, thank you very much. $\endgroup$ – user296355 Jan 10 '16 at 21:58

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