Problem 1

Exercise 10.1 (Connectedness: the plane & a lexicographic order topology TODO)  

1. Show that open, connected subsets of the plane are path-connected.
2. Show that $X=[0,1]\times[0,1]$ in the lexicographic order topology is not path-connected.

Proof. For part (a), let $x\in U$ be a fixed base point. Define $x\sim y$ if and only if there exists a continuous function $\gamma : [0,1] \to U$ such that $\gamma(0)=x$ and $\gamma(1)=y$. Consider the equivalence class of the base point.

$$V = \{y\in U \;\vert\; x\sim y\}$$ (10.1)

We show $V$ is open by constructing a neighborhood for any point. Let $y\in V$. Select a neighborhood $B_\delta(y)\cap U\subseteq U$. Then this same ball is also interior to $V$, for if $y'\in B_\delta(y)\cap V$, the convexity of the ball implies $y'\sim y$. The selection of $y$ precisely states $y\sim x$, so the transitivity of the relation implies $y' \sim x$. Therefore, $y'\in V$, so that $V$ is an open set. Writing

$$U = V\cup (U\setminus V)$$ (10.2)

and applying the connectedness of $U$ implies $V=U$, so that $U$ is path-connected.

The above proof could have also been performed in a constructive way as follows. Select a countable collection of rectangles such that

$$U = \bigcup_{i=1}^\infty R_i$$ (10.3)

We can explicitly construct the path component of $U$ in a tree based fashion, walking through the path-connected neighbors (sufficiently induced by their non-trivial intersection) starting at a base rectangle. Let $R_\ell \sim R_k$ if and only if $R_\ell\cap R_k$. Define the base level of a tree by

$$T_1 = \{R_1\}$$ (10.4)

and proceed in an inductive fashion, walking the rectangles $R_i\sim R_j$ for some $R_j\in T_n$ which have not already been walked. Formally, this can be written

$$T_{n+1} = \{R_i \;\vert\; R_j\in T_n R_i\cap R_j \}\setminus \bigcup_{m=1}^n T_m.$$ (10.5)

To complete the procedure, consider the entire tree we just generated

$$T = \bigcup_{n=1}^\infty T_n.$$ (10.6)

Now we wish to argue that $T = \{R_1, R_2, \dots\}$. Consider the complement $S=\{R_1,R_2,\dots\}\setminus T = \{R^1, R^2, \dots\}$, which we now argue is empty. Rewriting $U$

$$U = \left(\bigcup_{k=1}^\infty R^k\right) \cup \left(\bigcup_{R_i\in T}^\infty R_i\right),$$ (10.7)

connectedness implies some $R^k\cap R_i$ is non-empty. By convexity of rectangles, this means $R^k\sim R_i$. Realizing $R_i\in T_m$ implies $R^k\in T_{m-1}$ or $R^k\in T_{m+1}$, so that $R^k\in T$, showing that $S$ is empty.

Therefore,

$$U = \bigcup_{U_i\in T} U_i$$ (10.8)

From the construction of $T$, there is an explicit path between any two points, namely from a point to the base to the other point.

TODO For part (b), recall that the lexicographic order is induced by relation $(a,b) < (c,d)$ if and only if $a<c$ or $a=c$ and $b<d$. Explicitly, the basic sets are

$$\mathcal{B} = \left\{ \{x \in [0,1]\times[0,1] \;\vert\; a < x < b\} \;\vert\; a,b\in[0,1]\times[0,1]\right\}$$ (10.9)

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