Given an embedded disk in $\mathbb{R}^n$, is there always another disk which intersects it nontrivially in a disk?

Question

We call an open subset $D\subset X$ of a manifold $X$ an embedded disk, if there exists a homeomorphism $D\cong \mathbb{R}^n$.

The precise formulation of the question in the title is as follows:

Let $X=\mathbb{R}^n$ and fix a point $p\in X$. Let $p\in U\subsetneq X$ be an embedded disk containing the point. Can one always find another embedded disk $V\subsetneq X$ which intersects $U$ nontrivially in a disk away from $p$, i.e.:

• $p\notin V$ and $V\not\subseteq U$;
• the intersection $U\cap V$ is itself an embedded disk.

Some considerations/remarks:

• I would be somewhat happy with a version of this statement using other reasonable notions of "embedded disk" $D\subset X$. For example one could require the existence of a diffeomorphism $D\cong \mathrm{R}^n$ (rather than a homemorphism). Even stronger, one could require this diffeomorphism/homeomorphism to be (smoothly) isotopic to the given inclusion $D\hookrightarrow X=\mathbb{R}^n$.
• Any manifold $X$ for which the question has a positive answer has to be contractible, because it can be obtained inductively from a disk by attachments along homotopy equivalences of the form $V\cap U\hookrightarrow V$. This means that any proof needs to use the global topology of the ambient manifold $X=\mathbb{R}^n$ somehow (or at least its homotopy type).
• An explicit counterexample without the assumption $X=\mathbb{R}^n$ is given by the open annulus $X=A=\{x\in \mathbb{R^2}\mid 1<|x|<2\}$ and the embedded disk $U=A\setminus\{(0,x)\mid 1<x<2\}$ obtained by removing a radius (here $p$ is any point in $U$).
• In general, one cannot extend the embedding $\mathbb{R}^n\cong U\hookrightarrow X=\mathbb{R}^n$ to an embedding of the closed disk $\mathbb{R}^n\subset\overline{\mathbb{R}^n}$. For example, consider the example where $U=\mathbb{R}^2\setminus\{(0,x)\mid x\geq 1\}$ is obtained by removing a closed ray.

Answer 1

Here is a proof in the 2-dimensional case. It does not generalize in higher dimensions and more ideas would be needed. Reading R.Berlanga "A mapping theorem for topological sigma-compact manifolds", Compositio Math, 1987, vol. 63, 209-216 might be useful.

Let me start with some generalities. I will be considering the 2-sphere $S^2$ as the 1-point compactification of the Euclidean plane $E^2$, $S^2= E^2\cup \{\infty\}$. A compact metrizable connected space is called a continuum. A proper compact subset $K\subset S^2$ has complement homeomorphic to $E^2$ if and only if $K$ is a continuum satisfying $H^1(K)=0$ (here and in what follows all cohomology groups are Chech with integer coefficients). The latter condition is equivalent to connectedness of the complement of $K$ in $S^2$. In particular, if $K\subset E^2$ is compact, then its complement in $E^2$ is not homeomorphic to $E^2$. For a subset $C\subset E^2$, I will use the notation $\hat{C}=C\cup \{\infty\}$.

The following lemma is elementary, I omit the proof:

Lemma. Suppose that $K\subset S^2$ is a continuum whose complement is the union of components $U_i, i\in I$. Then for every $J\subset I$, the union $$ \tilde{K}= K\cup \bigcup_{j\in J} U_j $$ is again a continuum.

Now, I will prove that your question has positive answer in dimension 2.

Suppose that $C\subset E^2$ is a closed nonempty subset whose complement $U= E^2- C$ is homeomorphic to $E^2$. Then $\hat{C}=C\cup \{\infty\}$ is a continuum. There are two cases to consider:

1. $C$ has empty interior.

Pick an open subset $U\subset \hat{C}$ containing $\infty$. Then, by the boundary bumping (see e.g. Corollary 5.5 on page 74 in Nadler's book "Continuum Theory"), there exists a subcontinuum $\hat{C}'\subset W\subset \hat{C}$ containing $\infty$. Since $\hat{C}$ has empty interior and connected complement, $\hat{C}'$ also has connected complement. In particular, $E^2 - C'$ is homeomorphic to $E^2$. I will be taking $W$ which is a proper open subset $\hat{C}$; then $C'\ne C$. Pick any point $p\in E^2 -C$. Then there exists a closed line interval $[p,q]\subset E^2$ such that $[p,q]\cap C'=\{q\}$. Thus, $[p,q]\cup \hat{C}'$ is homotopy-equivalent to $\hat{C}'$ and, hence, is acyclic. In particular, the complement $[p,q]\cup \hat{C}'$ in $S^2$ is homeomorphic to $E^2$. We obtain an open subset $$V:=U''=E^2 - ([p,q]\cup {C}')$$ whose intersection with $U=E^2 - C$ equals $U'= E^2 - (C\cup [p,q])$. By the above observations, complements to both $U', U''$ in $E^2$ are homeomorphic to $E^2$. Moreover, $U''$ misses the given point $p\in U$ and is not contained in $U$. (If it were contained in $U$ then $U'=U$ which is, of course, false.) Thus, in Case 1, we obtained the property you asked for.

2. $C$ has nonempty interior. The proof in this case is very similar to Case 1, except the complement to $\hat{C}'$ may be disconnected. Let $U_i, i\in I$, denote those complementary components of $C'$ in $E^2$ which are disjoint from $U$. Then, by the Lemma, $\tilde{C}'$, the union of $C'$ with all $U_i, i\in I$, is again a continuum. Its complement, $U'$, is connected. Hence, we proceed as in Case 1, replacing $C'$ with $\tilde{C}'$.