Prove that if $E$ is open and pathwise connected then $E$ is polygonally connected.

Question

For $x,y ∈ \mathbb{R}^d$, let $L(x,y) = \{(1−t)x+ty | 0 ≤ t ≤ 1\}$. A nonempty subset $E ⊆ \mathbb{R}^d$ is called polygonally connected if for any $x,y ∈ E$ there exists $x = x_0, x_1, ..., x_n =y ∈ E$ such that $L(x_{j−1},x_j)⊆E$ for all $j=1,...,n.$

Prove that if $E$ is open and pathwise connected then $E$ is polygonally connected.

My attempt: E is pathwise connected, then there exists a $v:[0,1]\to E$, for all $n \in \mathbb{N}$, $v(1/n)$ are on the path. Then $L(v(1/a),v(1/{a+1}))$ are under $E$.

I feel like I can't say the last sentence, can you give me some hints or point me to the right direction if I'm heading towards the wrong way. Thanks a lot in advance.

Answer 1

Take $x \in E$. Let $A \subseteq E$ be the set of all points $y \in E$ such that $x,y$ are polygonally connected through $E$. Note $A$ is nonempty, since it contains $x$. Also, $A$ is open in $E$ since for any $y_1 \in A$, we can take a small neighborhood $U$ around $y_1$ such that $U \subseteq E$, and then for any $y_2 \in U$, we can trivially polygonally connect $y_1$ to $y_2$. Finally, $A$ is closed in $E$, since if $(x_n)_n \in A$, $x_n \to x \in E$, then we can find an open set $U \ni x$ contained in $E$; and since some $x_n$ is in $U$, we can polygonally connect $x_n$ to $x$ to see that $x \in A$. Since $E$ is connected, we see $A = E$, as desired.

Answer 2

You are right, the last step doesn't seem right. I am not sure how to make that work. I don't see how you are using that $E$ is open either.

Probably the best(imho) way to go about this problem is the following. The interval $[0,1]$ is compact, so th image $v([0,1])$ is also compact. Furthermore, $v([0,1])\subseteq E$, so for each $x\in [0,1]$ there exists an $\varepsilon_x>0$ with $B(v(x),\varepsilon_x)\subseteq E$. Now use compactness to get finitely many such balls. Since $[0,1]$ is connected so is its image, so these balls must intersect each other sufficiently often that you can connect the dots.

Answer 3

Suppose that $E$ is pathwise connected and open. Consider $x, y \in E$. Because $E$ is pathwise connected, there exists a path $p : [0,1] \rightarrow E$, continuous, such that $p(0)=x$ and $p(1)=y$.

Now because $E$ is open, for all $t \in [0,1]$, there exists an open ball $B_t$ around $p(t)$ and contained in $E$. Obviously the family $(B_t)_{t \in [0,1]}$ covers $p([0,1])$, which is compact, so you can extract a finite cover $B_1$, ..., $B_n$ of $p([0,1])$ by open balls.

The union of the balls $B_i$ is connected (because it contains $p([0,1])$), so you can suppose that $B_1$ contains $x$, $B_n$ contains $y$, and for each $i=1, ..., n-1$, $B_i \cap B_{i+1} \neq \emptyset$. Choose for each $i = 1, ..., n-1$ a point $z_i \in B_i \cap B_{i+1}$. The paths $L(z_i, z_{i+1})$ are contained in $B_i$ (so also in $E$) for $i=2, ..., n-2$ ; and the paths $L(x,z_1)$ and $L(z_{n-1}, y)$ are contained respectively in $B_1$ and $B_n$ so also in $E$.

The sequence of points $x, z_1, ..., z_n, y$ shows that $E$ is polygonally connected.

Answer 4

Fix $p \in E$.

Let

$$F=\{x \in E: \exists n \ge 0 \exists x_0,\ldots,x_n: (x_0=p) \land (x_n =x) \land \forall 1\le i \le n: L(x_{i-1},x_i) \subseteq E\}$$

Then $F$ is non-empty (e.g. $p \in F$), and as open balls in $\mathbb{R}^d$ are convex, we can easily check that $F$ is open and also closed in $E$ (show e.g. that a point in its closure can be connected to $p$ this way too). As $E$ is path-connected thus connected we see that $E=F$ and you are done.