# Closure and boundary of a bounded $\Lambda^\infty$-set

Let $\Lambda = (\Lambda_n)_{n \in \NN}$ be a system of collections $\Lambda_n$ of subsets of $\RR^n$ satisfying Axioms $(\Lambda 1)$–$(\Lambda 7)$.

The goal of this post is to describe the topological closure and boundary of a bounded $\Lambda^\infty$-set.

To fix notations below, let $X \subseteq \RR^n$ be a bounded, basic $\Lambda^\infty$-set. Let $W \subseteq \RR^{k+n+l}$ be a $\Lambda$-set such that $X$ is obtained from $W$, say $$X = \bigcup_{i \in \NN} X(i)$$ with $X(1) \subseteq X(2) \subseteq \cdots$ and, for each $i,j \in \NN$, let $z(i,j) \in \RR^k$ be such that each set $X(i,j):= \Pi_nW_{z(i,j)}$ is bounded and $$X(i) = \lim_j X(i,j).$$

First, the easy one:

Lemma 1
$\cl(X)$ is a basic $\Lambda^\infty$-set obtained from $W$.

Proof. Since the sequence of all $X(i)$ is increasing and $X$ is bounded, we have $$\cl(X) = \lim_i X(i).$$ The lemma therefore follows from Fact 4. $\qed$

The case of the boundary is a bit harder to deal with: there is no obvious representation of $\bd(X)$ as a $\Lambda^\infty$-set. Instead, we show:

Proposition
There is a sparse, closed $\Lambda^\infty$-set that contains $\bd(X)$.

Proof.
Note first that we may assume that $W$ is closed and each fiber $W_z$ is compact.
(see how)

By $(\Lambda 5)$, there exist $l’ \ge l$ and a closed $\Lambda$-set $W’ \subseteq \RR^{k+n+l’}$ such that $$W = \Pi_{k+n+l} W’;$$ note that $X$ is a $\Lambda^\infty$-set obtained from $W’$ as well. Next, we define $$V:= \set{(z,t,x,y’):\ (z,x,y’) \in W’ \text{ and } \left|(x,y’)\right| \le t}.$$ Note that, for $z \in \RR^k$, each fiber $V_{z,t}$ with $t \in \RR$ is compact, and that $$W’_z = \bigcup_{t>0} V_{z,t}$$ and $$\Pi_n W_z = \Pi_n W’_z = \bigcup_{t>0} \Pi_n V_{z,t}.$$ Since $V_{z,t} \subseteq V_{z,s}$ whenever $s \ge t$, and since $X(i,j)$ is bounded for each $i$ and $j$, it follows that $$\cl X(i,j) = \lim_{t \in \NN, t \to \infty} \Pi_nV_{z(i,j),t}.$$ It follows from Fact 4 that $X$ is a $\Lambda^\infty$-set obtained from $V$ as well.

In this situation, we use $(\Lambda 7)$ to write $W$ as a finite union of $\Lambda$-sets $W^1, \dots, W^q$ that are manifolds in standard position. For $p=1, \dots, q$, we let $n_p \le n$ be the constant rank of $\Pi_n\rest{W^p_z}$ for $z \in \Pi_k\left(W^p\right)$, which is independent of $z$ by assumption. We set $$I:= \set{p:\ n_p < n}.$$ Exercise
Show that $I \ne \emptyset$, and that $\Pi_n W^p_z$ is meager for $p \in I$ and $z \in \RR^k$.
[Hint: use the remark following the definition of “manifold in standard position” in this post.]

For each $p$ and $i$, since $W^p \subseteq W$ and the sets $X(i,j)$ converge to $X(i)$, the sequence of all $$X^p(i,j):= \Pi_nW^p_{z(i,j)}$$ has a Hausdorff limit $X^p(i) \subseteq X(i)$. Moreover, since $X$ is bounded, the sequence of all $X^p(i)$ has a Hausdorff limit $X^p \subseteq \cl(X)$. Passing to a subsequence if necessary, we may assume, for each $p$, that $$X^p(i) = \lim_j X^p(i,j) \quad\text{for each } i$$ and $$X^p = \lim_i X^p(i).$$

Exercise
Show that each $X^p$ is a closed, basic $\Lambda^\infty$-set obtained from $W^p$, and that $$\cl(X) = X^1 \cup \cdots \cup X^p.$$

Finally, we define $$Y:= \bigcup_{p \in I} X^p;$$ the set $Y$ is a closed $\Lambda^\infty$-set by the previous exercise. Moreover, by the previous two exercises and this theorem, the set $Y$ is sparse. So it remains to prove:

Claim
The set $Y$ contains $\bd(X)$.

To see this note that, since $X$ is the union of the increasing sequence of compact sets $X(i)$, every point $x \in \bd(X)$ is the limit of some sequence of points $x_i \in \bd(X(i))$. So we fix $i \in \NN$ and show that $$\tag{\ast}\bd(X(i)) \subseteq \bigcup_{p \in I} X^p(i).$$

Let $a \in \bd(X(i))$ and $r\gt 0$; to show $(\ast)$, it now suffices to show that there exist $p \in I$ and $j \in \NN$ such that $$\tag{\ast\ast}B(a,r) \cap X^p(i,j) \ne \emptyset.$$ Let $b \in B(a,r) \setminus X(i)$; since the sequence of projections $X(i,j)$ converges to $X(i)$, there exists $j_0 \in \NN$ such that $b \notin X(i,j)$ for $j \ge j_0$. Since $a \in X(i)$, it follows from Fact 3 that there are $j\ge j_0$ and $c \in X(i,j)$ such that $c \in B(a,r)$.

But the segment $[b,c]$ is contained in $B(a,r)$, and since $X(i,j) \cap [b,c]$ is nonempty, compact and does not contain $b$, it contains some $d \in [b,c]$ that is closest to $b$.

However, the sets $X^p(i,j)$ with $p \notin I$ are open; so $d$ has to belong to some $X^p(i,j)$ with $p \in I$, and $(\ast\ast)$ is proved. $\qed$

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