Proof that plane has small inductive dimension 2?

Question

By definition, the small inductive dimension $\operatorname{ind}(\emptyset) = -1$ and, recursively, the small inductive dimension $\operatorname{ind}(X)$ of a nonempty topological space $X$ is the least integer $n \geq 0$ such that each point $x$ has a local base (i.e., a neighborhood base) of open sets $V$ such that $\operatorname{ind}(\operatorname{\partial} V) \leq n - 1$, where $\operatorname{\partial} V$ denotes the boundary of $V$.

(Some versions of the definition require that each neighborhood $U$ of $x$ contain such a $V$ with $\operatorname{cls} V \subset U$. However, that requirement here is redundant since the spaces involved are regular.)

That $\operatorname{ind}(\mathbb{R}) = 1$ is easy to see: $\mathbb{R} \neq \emptyset$; and each point in $\mathbb{R}$ has arbitrarily small neighborhoods of the form $V = (a, b)$, and $\partial\,V$ is the two-point discrete space $\{a, b\}$, which is 0-dimensional.

But what about $\mathbb{R}^2$: is there an elementary proof that $\operatorname{ind} (\mathbb{R}^2) = 2$?

Evidently $\operatorname{ind}(\mathbb{R}^2) \leq 2$, since each point has arbitrarily small neighborhoods that are open disks, and the boundary of such a disk has small inductive dimension $1$. Moreover, $\operatorname{ind}(\mathbb{R}^2) \neq 0$, since the plane is connected; and of course $\operatorname{ind}(\mathbb{R}^2) \neq -1$.

So the thing that remains to prove is that $\operatorname{ind} (\mathbb{R}^2) \neq 1$.

I know there are not-so-elementary proofs that $\operatorname{ind} (\mathbb{R}^n) = n$, but I'm looking for an elementary proof just in the case $n = 2$.

Answer 1

In my book

Edgar, Gerald A., Measure, topology, and fractal geometry, Undergraduate Texts in Mathematics. New York etc.: Springer-Verlag. ix, 230 p. DM 58.00/hbk (1990). ZBL0727.28003.

I included what I considered to be the most elementary proof of this fact, Theorem 3.3.4. [It's about 4 pages.] This is done first talking about degree (mod 2) of a map from the circle to itself; then Brouwer's fixed point theorem for the disk; then the topological dimension of the plane.

In the first edition of the book, where I do inductive dimension first, this is done with inductive dimension. In the second edition of the book I switched to the more common arrangement, doing covering dimension first, so this is done with covering dimension.

Answer 2

It suffices to show that for every bounded open set $V$ on the plane, its boundary $\partial V$ has a connected subset that has more than one point. Let $V_1$ be a connected component of $V$. Then $K=\overline{V_1}$ is compact and connected. Its complement $K^c$ contains exactly one unbounded connected component $D$. For this domain $D$, its complement $D^c$ consists of $K$ together with the bounded domains in $K^c$, so it is elementary to verify that $D^c$ is connected. Thus the union $D\cup \{\infty\}$ is simply connected on the Riemann sphere, see [2]. Therefore the boundary $\partial D$ is connected, see e.g. [1]. Certainly $\partial D$ contains more than 1 point. But $\partial D \subset \partial K =\partial V_1 \subset \partial V$.

[1] Is it true that a boundary of a simply connected and bounded set is connected in $\mathbb{C}$?

[2] Complement is connected iff Connected components are Simply Connected

Answer 3

This answer is supposed to be a complement to the answer offered by Yuval Perez, with the following observations:

1. I am not convinced of the proof given in the link Is it true that a boundary of a simply connected and bounded set is connected in $\mathbb{C}$?, so my answer essentially attempts to fix the part of Yuval's answer that uses that link.

2. As I mentioned in a comment below Yuval's answer, I am not entirely comfortable with the claim that the set denoted $D$ in Yuval's answer is indeed simply connected (after $\infty $ is added to it). Unfortunately I have no easy fix for this, hence my own answer is dependent on this point.

This said, as noted by Yuval, it suffices to prove the following:

Lemma. If $U$ is a nonempty, open, bounded, simply connected subset of ${\mathbb R}^2$, then the boundary $\partial U$ is connected.

Proof. Identifying ${\mathbb R}^2$ with the complex plane, and using the Riemann mapping theorem, there is a bi-holomorphic map $f$ from the open unit disk $\mathbb D$ onto $U$.

For each real number $r$ in the interval $(0,1)$, consider the subset of $U$ defined by $$ C_r= \{f(z): |z|=r\}. $$

Roughly speaking, we will prove that the $C_r$ converge to $\partial U$, and since each $C_r$ is connected, so will be $\partial U$. The details are as follows:

Fixing $\varepsilon >0$, denote by $V_\varepsilon $ the open set $$ V_\varepsilon :=\{x\in U: \text{dist}(x, \partial U)<\varepsilon \}. \tag {1} $$ We then claim that there is some $r_0\in (0,1)$ such that, for every $r$ in $(0,1)$, with $r>r_0$, one has that $$ C_r\subseteq V_\varepsilon . $$

To prove the claim, observe that $$ K_\varepsilon := U\setminus V_\varepsilon = \{x\in U: \text{dist}(x, \partial U)\geq \varepsilon \} $$ is a compact set (Reason: every sequence in $K_\varepsilon $ has a sub-sequence wich converges in $\bar U$, say with limit $x$, because $\bar U$ is compact. By continuity $\text{dist}(x, \partial U)\geq \varepsilon $, so $x$ is not in $\partial U$, meaning that $x\in U$).

The collection of sets $$ D_r:= \{f(z) : |z|<r\}, $$ for $r$ ranging in $(0,1)$, clearly forms a cover for $U$, and in particular also for $K_\varepsilon $. Since the $D_r$ are increasing, by compactness there is some $r_0$ such that $K_\varepsilon \subseteq D_{r_0}$, and hence also $K_\varepsilon \subseteq D_r$, for every $r\geq r_0$. It follows that $$ C_r \subseteq U\setminus D_r \subseteq U \setminus K_\varepsilon = V_\varepsilon , $$ as desired.

Assuming by contradiction that $\partial U$ is disconnected, write $\partial U=B_1\sqcup B_2$, where $B_1$ and $B_2$ are nonempty, closed sets, "$\sqcup$" standing for disjoint union.

Since both $B_1$ and $B_2$ are closed and bounded, they are compact and hence their distance is strictly positive. In symbols $$ d:= \inf\{|x-y|: x\in B_1, \ y\in B_2\} >0. $$ We then consider, for every $i=1,2$, the open set $$ W_i=\{x\in U: \text{dist}(x, B_i) < d/2\}. $$ Evidently $W_1\cap W_2=\emptyset$, while $W_1\cup W_2=V_{d/2}$ (as defined in (1)).

Taking $r_0$, as above, for the choice of $\varepsilon =d/2$, we conclude that for every $r$ in the interval $(r_0, 1)$, one has that $$ C_r\subseteq W_1\cup W_2. $$ Observing that $C_r$ is connected, we deduce that either $C_r\subseteq W_1$, or $C_r\subseteq W_2$.

However, it is a simple matter to show that, for every $i$, the set $$ J_i:= \{r\in (r_0, 1): C_r\subseteq W_i\} $$ is open. Since $(r_0, 1)= J_1\sqcup J_2$, we have by connectedness that either $J_1$ or $J_2$ coincides with $(r_0,1)$. We then suppose, without loss of generality that $J_1=(r_0, 1)$, which is to say that $$ C_r\subseteq W_1,\quad \forall r\in (r_0,1).\tag {2} $$

Pick some point $x$ in $B_2$, and let $\{x_n\}_n$ be a sequence in $U$, converging to $x$. Assuming, as we may, that $|x_n-x|<d/2$, we necessarily have that $x_n\in W_2$.

Write $x_n=f(z_n)$, where $z_n$ lies in the open unit disk $\mathbb D$. By passing to a subsequence, we may assume that $\{z_n\}_n$ converges to some point $z\in \bar {\mathbb D}$. Clearly $z$ cannot lie in ${\mathbb D}$, because otherwise $$ x = \lim_n x_n = \lim_n f(z_n) = f(z) \in U, $$ but we know that $x\in B_2\subseteq \partial U$.

It follows that $|z|=1$, so $\lim_n|z_n|=1$, and hence there is some $n_0$ such that $n\geq n_0$ entails $|z_n|>r_0$. This implies that, for $n\geq n_0$, $$ x_n=f(z_n) \in C_{|z_n|} \subseteq W_1, $$ by (2), contradicting the fact that $x_n$ lies in $W_2$. QED