Let $D \subset \Bbb R^n$ be an open connected set. I would like to exhibit an increasing sequence of compact connected subsets of $D$ converging to $D$.

For example, for a ball we might take a sequence of closed ball inside it of radius $r-\frac 1n$.

User SteamyRoot showed that the usual closed exhaustion that generalizes the above example, given by $D_n = \{ x\in D: \mathrm{dist}(x, \partial D) \geq \frac 1n \}$ is not always connected (by considering the shape of eye-glasses), even before modifying it to be bounded. Considering an infinite sequence of glasses connected to one another in a row that are smaller and smaller shows that this will not be connected no matter how small we choose $n$.

We also cannot do this by taking the union of all closed squares of side length $2^{-n}$ contained entirely within $D$. This is again not connected because of the same counterexample (infinite glasses), even though this example does have such an exhaustion.

  • $\begingroup$ You could also connect the components of $D_n$ by a path which wouldn't change as $n$ increases. $\endgroup$ – Robert Wolfe Sep 1 '17 at 20:25

Fix a point $x_0 \in D$. For every $A$ subset of $D$ containing $x_0$ denote by $A'$ the component of $A$ containing $x_0$. Assume now we have $D_m$ an increasing sequence of open subsets of $D$ containing $x_0$ and with union $D$. Then $D_m'$ (the $x_0$ components) also cover $D$. Indeed, consider $x \in D$. There exists a path $\gamma$ from $x_0$ to $x$ contained in $D$. Since the $D_n$'s cover $D$ and $\gamma$ is compact there exists $n$ so that $\gamma \subset D_m$. Then $x\in D_m'$.

Take now any increasing sequence of compacts $K_m$ whose interiors cover $D$ ( for instance $K_m= \{ x \in \mathbb{R}^n \ | \ ||x|| \le m \textrm { and } d(x, \mathbb{R}^n \backslash D ) \ge \frac{1}{m} \}$). Then $L_m = K_m'$ will be an exhaustion of $D$ with connected compacts.

$\bf{Added:}$ In the construction above, we can also take $L_m$ to the the closure of the $x_0$ connected component of $\mathring{K_m}$ ( the interior of $K_m$). The advantage is that the interior of $L_m$ is connected and its closure is $L_m$, so $L_m$ is better behaved.

In general, if $L$ is a compact subset of the open set $D$, consider a covering of the full space with a lattice of $\epsilon$ cubes so that the diagonal of the cubes $\sqrt{n} \cdot \epsilon < d(L, \mathbb{R}^n \backslash D)$. Consider $\tilde L$ to be the union of all the cubes that intersect $L$. Then $\tilde L$ is a compact and $L \subset \tilde L \subset D$. If, moreover, $L$ is connected then $\tilde L$ is connected. If $L$ has connected interior $\mathring{L}$ and is the closure of its interior, then the same holds for $\tilde{L}$. So we can get pretty well behaved compact exhaustions.

  • $\begingroup$ Concretely, if $\gamma\colon [0,1]\to D$ is a path from $x_0$ to $y$, then the continuous function $t\mapsto \|\gamma(t)\|$ attains its maximum $R$ and the continuous function $t\mapsto d(\|\gamma(t)\|,\Bbb R^n\setminus D)$ attains its (positive) minimum $r$. Then $y\in L_m$ for all $m>\max\{R,1/r\}$. $\endgroup$ – Hagen von Eitzen Sep 2 '17 at 9:41

Yes. You can do this by fiddling with an arbitrary sequence of compact sets exhausting the space. In particular, let us first choose any increasing sequence $C_1\subseteq C_2\subseteq C_3\subseteq\ldots$ of compact sets with $\bigcup_i C_i=D$.

First, we wish to fatten the set to avoid issues that would arise if we studied a set like $\{1/n:n\in\mathbb Z\}\cup \{0\}$ which is compact but has infinitely many connected components. To this end for each $i$, define a set $$C'_i=\{x\in D:d(x,\partial D)\geq d(x,C_i),\,d(x,0)\leq \max_{p\in C_i}d(p,0)+1\}.$$ This set remains compact, being closed and bounded by definition. Observe that $C_i\subseteq C'_i$ and that $C'_i\subseteq C'_{i+1}$.

First, note that if $y\in C'_i$, then there is some $x\in C_i$ minimizing $d(y,x)$ by compactness. It may be seen that the line segment from $y$ to $x$ is contained within $C'_i$. Moreover, one may check that every $x\in C_i$ is in the interior of $C'_i$.

In particular, we find the following statement:

$C_i$ is covered by the interiors of the connected components of $C'_i$.

Because the connected components are by definition disjoint and each contains an element of $C_i$ in its interior, this cover has no proper subcovers. By compactness, it is therefore finite and hence $C'_i$ has finitely many connected components.

Now, we may define a set $C''_i$ by choosing some set of paths $\gamma$ in $D$ that connect the components to each other. By choosing these paths consistently, we may construct an increasing sequence $C''_i$ of compact sets exhausting the space.


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