Can the infinite jungle gym surface be expressed by an exhaustion of compact surfaces with one boundary component?

Question

Is it possible to write the infinite jungle gym surface as the increasing union of compact surfaces whose boundaries consist of only one simple closed curve? I believe that this is true. This surface has only one end. Therefore the result follows from the existence of canonical exhaustions (made of compact surfaces whose components of the complement have one boundary component) (as explained in Ahlfors-Sario book). If anyone could help me, I would like to "see" such a curve. Thanks.

Answer 1

It'll be quite hard to see in fine detail, but here's a somewhat course visualization.

Let $J$ be the infinite jungle gym surface.

Start with the sphere $S_r$ of arbitrarily large radius $r>0$ centered on the origin $\mathcal O$, which is the boundary of the ball $B_r$. You can perturb $r$ by an arbitrarily small value so that $S_r$ is transverse to $J$, and hence $J \cap S_r$ consists of a union of circles. The complement $J - S_r$ has a unique unbounded component, coinciding with $J - B_r$. To prove this one has to show that for each point $x_0 \in J$ there is a continuous path $x_t$, $t \in [0,\infty) \to J$ which starts at $x_0$, such that the distance $d(\mathcal O,x_t)$ is nondecreasing and diverges to $+\infty$.

The intersection $J \cap S_r$ consists of a zillion circles, which I'll enumerate as $C_i$, $i=1,...,$ (a zillion). What "a zillion" really means here is that the number of these circles grows quadratically as a function of $r$, being asymptotically comparable to the area of $S_r$ which is equal to $4 \pi r^2$.

Now start connecting them up by a network of pairwise disjoint paths in $J - B_r$ (this where is an explicit "visualization" really begins to break down). To be precise, take a properly embedded arc $\gamma_{1} \subset J$ with one endpoint on each of $C_1$ and $C_2$. This arc $\gamma_1$ does not separate $J$, because $C_1 \cup \gamma_1 \cup C_2$ has a regular neighborhood in $J$ whose boundary is a single circle. Thus we can take a properly embedded arc $\gamma_2 \subset J$ with one endpoint on each of $C_2$ and $C_3$ and disjoint from $\gamma_1$. Again the union $\gamma_1 \cup \gamma_2$ does not separate $J$, because $C_1 \cup \gamma_1 \cup C_2 \cup \gamma_2 \cup C_3$ also has a regular neighborhood in $J$ whose boundary is a single circle.

Continuing inductively, we obtain a system of circles and arcs $$C_1 \cup \gamma_1 \cup C_2 \cup \gamma_2 \cup C_3 \cup \cdots \cup C_{\text{a zillion}} $$ which has a regular neighborhood in $J$ whose boundary is the single circle that you have asked for.

What you can visualize here is a long string of disjoint little circles in the plane, of equal radius, strung out with their centers on the $x$ axis, and connected up by arcs of the $x$-axis. Also you should visualize a big circle which encloses this whole string of centers and arcs. The union of the little circles is the boundary of $J$, the portion of the plane outside these little circles can be thought of as embedded in $J$ (that's the final regular neighborhood), and the final circle you want is any circle in the plane whose inside contains all of the little circles.