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I am an undergraduate in mathematics, self-studying complex analysis. In Bak, Newman - Complex Analysis, there is the following definition:

An open connected set $D subset mathbb{C}$ is simply connected if its complement is "connected within $epsilon$ to $infty$". That is, for any $z_0 in mathbb{C}-D$ and $epsilon>0$, there exists a continuous curve $gamma :[0,infty) rightarrow mathbb{C}$ such that

(a)$d(gamma(t), mathbb{C}-D) <epsilon $ for all $t geq 0$,

(b) $gamma(0)= z_0$,

(c) $lim _{trightarrow infty} gamma(t) = infty$.

I blowed up my mind, since in my topology class I learned the following definition:

A simply connected space is a path-connected space whose fundamental group vanish.

How can I prove that these two seemingly very different definitions are actually equivalent(if it is indeed equivalent)?

To be honest, the definition in Bak and Newman is the wort definition of simple connectivity I had ever seen. I will prove its equivalence to the traditional definition below. The proof is not short and is definitely outside of the traditional complex analysis syllabus.

In what follows, I identify the 1-point compactification of $R^n$, $R^ncup {infty}$ with the sphere $S^n$.

I will use the notation $B(c,r)={xin R^n: |x-c|<r}$ for the open $r$-ball centered at $c$.

Definition. Given a closed subset $Asubset R^n$ and $epsilon>0$, define the $epsilon$-neighborhood of $A$ in $R^n$ as
$$
N_epsilon(A):= {xin R^n: exists yin A, |x-y|<epsilon} = bigcup_{ain A} B(a,r).
$$
Then
$$
bar{N}_epsilon(A):= {xin R^n: exists yin A, |x-y|leepsilon}
$$
is the closure of $N_epsilon(A)$ in $R^n$.

Definition. I will say that a subset $Asubset R^n$ satisfies the BN (for Bak and Newman) property if for every $epsilon>0$ and every $ain A$, there exists a continuous map $q: [0,infty)to N_epsilon(A)$ with $q(0)=a$ and
$$
lim_{ttoinfty} q(t)=infty,
$$
i.e.
$$
lim_{ttoinfty} |q(t)|=infty.
$$

Our goal is to prove

Theorem 1. The following are equivalent for open connected subsets $Usubset R^2$:

i. $U$ is simply connected in the traditional sense, i.e. every continuous map $S^1to U$ extends to a continuous map $D^2to U$.

ii. The complement $A:= R^2-U$ satisfies the BN property.

The proof is somewhat tricky: The direction ii$Rightarrow$i is rather straightforward, but the converse will require some work.
I could not find any proof shorter than the one written below.

I will be using the following characterization of simply-connected open planar sets; see here or my argument here for a proof.

Theorem 2. The following are equivalent for a nonempty open connected subset $Usubset R^2$:

1. $U$ is simply connected in the traditional sense, i.e. every continuous map $S^1to U$ extends to a continuous map $D^2to U$.




2. $S^2-U$ is connected.

In view of this theorem, we will need to analyze closed subsets $Asubset R^2$ such that $hat{A}:= A cup{infty}$ is connected.
The proofs will go through in all dimensions; therefore, I will be considering subsets of $R^n$, $nge 1$. Note that a subset $Asubset R^n$ such that $hat{A}:= A cup{infty}$ is connected, is necessarily unbounded. Then $hat{A}$ is the closure of $A$ in $S^n$.

Lemma 1. For a closed subset $Asubset R^n$ the following are equivalent:

1. $hat{A}$ is connected.




2. For every open neighborhood $V$ of $A$ in $R^n$, every component $V_j$ of $V$ is either unbounded or
   is disjoint from $A$.




3. Every connected component of $A$ is unbounded.

Proof. Suppose that $hat{A}$ is connected but there exists an open neighborhood $V$ of $A$ in $R^n$ and a bounded
component $V_j$ of $V$ such that $V_jcap A=A_jne emptyset$. Then $A_j$ is clopen (both closed and open) in $A$ (in the subspace topology). Since $V_j$ is open in $S^n$, the subset $A_j$ is also open in $hat{A}$ (in the subspace topology). Since $V_j$ is bounded, so is $A_j$. Hence, $infty$ does not belong to the closure of $A_j$ in $S^n$, hence, $A_j$ is closed in $hat{A}$. This contradicts connectedness of $hat{A}$.

Conversely, suppose that 2 holds but $hat{A}$ is not connected. Then, since $hat{A}$ is compact, there exists a pair of disjoint open subsets $V_1, V_2subset S^n$ such that $inftyin V_1cap hat{A}$ and $V_2cap Ane emptyset$. But then $V_2$ is an open and bounded subset of $R^n$ whose intersection with $A$ is nonempty, contradicting 2.

Let us now prove equivalence of 2 and 3. Suppose that 3 holds. Consider an open neighborhood $V$ of $A$ in $R^n$
and a connected component $V_j$ of $V$. If $V_jcap Aneemptyset$ then $V_j$ contains a component $A_j$ of $A$. Since $A_j$ is unbounded, so is $V_j$.

Suppose that 2 holds, but $A$ contains a bounded connected component $A_1$. Consider the system of open $r$-neighborhoods
$N_r(A)$ for $r=1/i, iin {mathbb N}$. Then
$$
bigcap_{r} N_r(A)= bigcap_{r} bar{N}_r(A)=A.
$$
For each $i$ there exists a connected component $V_i$ of $N_{1/i}(A)$ containing the unbounded component $A_1$ of $A$. I assume that $A_1$ is contained in an open ball $B(R)$ of radius $R$. For each $i$, the intersection of $bar{V}_i$ (the closure of $V_i$ in $R^n$) with the boundary sphere $S(R)$ of the ball $B(R)$ is compact and nonempty. These intersections form a nested sequence of compact nonempty subsets:
$$
bar{V}_icap S(R) subset bar{V}_{i-1}cap S(R).
$$
Therefore, their intersection $K$ is nonempty and is disjoint from $A_1$. Hence,
$$
C:= bigcap_{i} bar{V}_i
$$
is strictly larger than $A_1$. At the same time, since $bar{V}_isubset bar{N}_{1/i}(A)$, $C$ is contained in $A$.
Lastly, since each $bar{V}_i$ is connected (as the closure of a connected subset $V_i$) their intersection $C$ is also connected.
Thus, $A_1$ is not a maximal connected subset of $A$, which is a contradiction. qed

Lemma 2. Suppose that $Asubset R^n$ is a closed subset which satisfies the BN property.
Then the set $hat{A}$ is connected.

Proof. Assume that $hat{A}= A cup{infty}$ is not connected. Then, in view of Lemma 1, there exists an open neighborhood $V$ of $A$ in $R^n$ and a bounded component $V_1$ of $V$ which has nonempty intersection $A_1=V_1cap A$ with $A$. Since
$A_1=bar{V}_1cap A$ is closed and bounded, it is compact; hence, there exists an $epsilon>0$ such that $N_epsilon(A_1)subset V_1$. For every path $p: [0,infty)to N_epsilon(A)$ such that $p(0)in A_1$, the image of $p$ is contained in $V_1$, i.e. is bounded. This contradicts the BN property. qed

The remainder is a proof of the converse to Lemma 2 (Lemma 7); the proof is substantially more difficult that the proof of the lemma.
First, I will need some technical results. Much of this is quite standard (in some areas of topology) but would be quite out of place in a course in Complex Analysis.

Definition. A continuous map $f: Xto Y$ between two metric spaces is called metrically proper if preimages of bounded subsets under $f$ are bounded.

Remark. If $X, Y$ satisfy the Heine-Borel property (closed and bounded subsets are compact), then a continuous map $f: Xto Y$
is metrically proper if and only if it is proper, i.e. preimages of compact sets are compact.

The following lemma is straightforward, I will omit the proof:

Lemma 3. A continuous map $q: Xto Y$ is metrically proper if and only if it sends sequences diverging to infinity to sequences diverging to infinity, i.e. for every sequence $x_iin X$ such that
$d(x_1, x_i)toinfty$, it follows that
$$
lim_{itoinfty} d(f(x_1), f(x_i))=infty.
$$

Corollary. A continuous map $q: [0,infty)to R^n$ is proper if and only if
$$
lim_{ttoinfty} |q(t)|=infty.
$$

I will be using metrically proper maps between graphs and $R^n$.

A graph $G$ is said to have finite valence if the number of edges incident to each vertex is finite (not necessarily uniformly bounded). Let $G$ be a connected graph. I will equip the graph $G$ with the graph-metric $d_G$ where every edge of $G$ has unit length (and is isometric to the init interval) and the distance between any two vertices is the length of the shortest edge-path between them. Thus, connected graphs become metric spaces.

Remark. A connected graph $G$ of finite valence, equipped with the graph metric $d_G$ as above, satisfies the Heine-Borel property.

Lemma 4. Suppose that $G$ is connected, has finite valence and is unbounded, i.e. has infinite diameter. Then $G$ contains a proper ray, i.e. there exists a (metrically) proper map $p: [0,infty)to G$ sending integers to vertices of $G$ and linear on every
interval $[n, n+1]$ (with the image equal to the edge spanned by $p(n), p(n+1)$).

Proof. Fix a vertex $uin G$ and consider a sequence of vertices $v_iin G$ such that $d_G(u,v_i)toinfty$. For each $i$
let $p_i$ denote a shortest path from $u$ to $v_i$. Then, using finite valence of $G$ prove that the sequence of paths $p_i$ has a convergent subsequence whose limit is
an injective map $p: [0,infty)to G$ sending integers to vertices of $G$ and linear on every
interval $[n, n+1]$. Then verify that $p$ is (metrically) proper. qed

Lemma 5. Fix a positive number $delta>0$.
Suppose that $G$ is a connected graph and $f: Gto R^n$ is a map which sends each edge of $G$ linearly
to a line segment. (I will call such maps "linear".) Assume that for any two distinct vertices $u, v$ of $G$,
$|f(u)- f(v)|ge delta$. Then the map $f$ is (metrically) proper.

Proof. First observe that for every bounded subset $Bsubset R^n$ the preimage $f^{-1}(B)$ contains only finite number of vertices
(because images of vertices of $G$ are $delta$-separated). From this conclude that $f$ is (metrically) proper. qed

Given a subset $Csubset R^n$ and a number $r>0$ define a graph $G=G(C,r)$, whose vertex set equals $C$ and two vertices $u, v$ are connected by an edge if and only if $|u-v|< r$. This graph comes equipped with a linear map $f: Gto R^n$ which sends
each vertex of $G$ to the corresponding point in $R^n$.

Suppose that $Asubset R^n$ is a closed connected subset; fix $r>0$. Define $Csubset A$ to be a
maximal subset of $A$ such that no two distinct points $x, yin A$ satisfy $|x-y|<r$ ($C$ is $r$-separated).
Then, in view of the maximality
of $C$,
$$
Asubset V= bigcup_{cin C} B(c,r).
$$
Define the graph $G=G(C,2r)$ and let $f: Gto R^n$ denote the associated linear map.
This graph $G$ has finite valence, since its vertices are $r$-separated in $R^n$.
For every edge $e=[u,v]$ of $G$,
$f(e)subset B(u,r)cup B(v,r)subset V$. Hence,
the image of $f$ is contained in the open neighborhood $V$ of $A$.

Lemma 6. The graph $G$ is connected.

Proof. Suppose that $G$ is disconnected. Then the vertex set of $G$ splits as the disjoint union $C_1sqcup C_2$ such that no vertex of $C_1$ is connected by an edge to $C_2$. Hence, the balls $B(c_1,r), B(c_2,r)$ are disjoint whenever $c_1in C_1, c_2in C_2$. Then
$$
Asubset bigcup_{cin C_1} B(c,r) sqcup bigcup_{cin C_2} B(c,r) =V,
$$
contradicting connectedness of $A$. qed

Notice that if $A$ is unbounded, $C$ is unbounded as well, hence, the graph $G$ has infinitely many vertices and, thus, is unbounded too.

At last we can prove

Lemma 7. Suppose that $Asubset R^n$ is a closed subset such that every connected component of $A$ is unbounded. Then $A$ satisfies the BN property.

Proof. Take $ain A$ and let $A_1$ be the connected component of $a$ in $A$. Then $A_1$ is a closed connected subset of
$R^n$. Take $r=epsilon$ and construct a maximal $r$-separated subset $Csubset A_1$ containing $a$ and
the connected graph $G=G(C,r)$ as above. By Lemma 4, $G$ contains a proper ray $p: [0,infty)to G$. By Lemma 5, the map $f: Gto R^n$ is proper, hence, the composition $q= fcirc p$ is a proper map $[0,infty)to R^n$. The image of this map is entirely contained in
$V=N_r(C)subset N_r(A)$. Hence, $A$ satisfies the BN property. qed

This concludes the proof of Theorem 1.