Given an arbitrary connected and compact set $S$ with holes in it, is there a name for the simply connected set formed by the union of $S$ and its holes?

For example, let $S = \{x\in \mathbb{R}^n\ |\ 0 < a \leq||x||^2_2 \leq b\}$, its hole is $H = \{x\in \mathbb{R}^n\ |\ ||x||^2_2 < a\}$. Is there a name for $S \cup H$?

If there is not a specific definition for it, can I just call this set "the union of $S$ and its holes", or is there a more precise way of describing it?

By a set with holes I refer to sets such as this, which is taken from the Wikipedia page on simply connected sets. Here is my attempt to properly define a hole (at least in $\mathbb{R}^n$), inspired by this answer to another question.

Given a connected and compact subset $S \subset \mathbb{R}^n$, its boundary $\partial S$ can be written as the finite union of connected and compact hypersurfaces $F_i \subset \mathbb{R}^{n-1}$, such that $\partial S = \bigcup_{i=1}^k F_i$. Each $F_i$ cuts $\mathbb{R}^n$ in exactly two pieces $A_i \subset \mathbb{R}^n$ and $B_i \subset \mathbb{R}^n$, such that the closure of $A_i$ is compact and the closure of $B_i$ is not compact. Then, we say that $A_i$ is inside $F_i$, and $B_i$ is outside. A hole $H_i$ of $S$ is any $A_i$ such that $A_i \cap S = \emptyset$. Then, the set of all holes is $H = \bigcup H_i$.

If $S \cup H = S$, then is said that $S$ has no holes. This condition is equivalent to $H=\emptyset$.

If $S \cup H \neq S$, then I want to know if $S \cup H$ has a proper name. This condition is equivalent to $H\neq\emptyset$, and $H$ is "filling the holes" of $S$.

  • $\begingroup$ What exactly is meant by a hole ? $\endgroup$ – Shailesh May 14 '16 at 2:49
  • $\begingroup$ I tried to edit my question to address yours. This is how I would describe the process of making a set with holes: take a simply connected set, remove subsets in its interior so that the resulting set is still connected. This subsets are now the so called holes of the resulting set. $\endgroup$ – jcmonteiro May 14 '16 at 3:11
  • $\begingroup$ Yes, but if all I've been given is a space, how do I find the "holes" to fill in? $\endgroup$ – user98602 May 14 '16 at 3:23
  • $\begingroup$ My point is: there is a space, it might have these "holes", but all I want to do is to name the union of the space and its "holes". Much like the union of a set and its boundary is called closure. $\endgroup$ – jcmonteiro May 14 '16 at 3:28
  • $\begingroup$ The set that I describe has these holes, so it cannot be simply connected by definition. I say that it is connected just because the proof I am building needs it to be. Otherwise, I understand that $S$ could even be not connected and the definitions given so far would still hold. $\endgroup$ – jcmonteiro May 14 '16 at 20:06

A concise definition of the object you describe is "the complement of the unbounded component of the complement of $S$".

In two dimensions, the term simply connected hull is sometimes used: for example, in a MathOverflow post and in other places you can find with a search.

But in dimensions $\ge 3$ "simply connected" would be misleading: filling in a torus $\mathbb{T}^2$ creates a solid torus, which is still not simply connected.

Another term in use is filled-in set, for example Filled Julia set. This one can be used in any dimension.

  • $\begingroup$ I am not a mathematician, so just to be sure. Would the definition "the complement of the unbounded component of the complement of $S$" still be valid in higher dimensions ($\geq 3$)? It seems to me that it would. Besides, calling these sets filled-in seems appropriate. $\endgroup$ – jcmonteiro May 14 '16 at 19:21
  • $\begingroup$ Yes, this works for filling in holes in all dimensions. $\endgroup$ – user147263 May 14 '16 at 19:37

I assume you mean a compact set in $\mathbb{R}^n$ (or some metric space) so that "bounded" makes sense. Also, that in your clarification you probably meant to say something like the following (but see the NOTE at the end)

"take a COMPACT simply connected set $T$ and remove ( open) subsets of the interior to get a compact connected subset $S$."

The goal being to describe $T$ in relation to $S.$

sandwich gives a good answer and from it we see that the key is that "hole" is something like "(a subsequently removed) bounded component of the interior of $T$ " Which becomes "bounded component of the complement of $S.$"

One could avoid double complements by saying "the union of $S$ and the bounded components of the complement of $S.$" That makes sense even in (some) situations which are not compact. For example

"Let $T$ be a (finite union of) CLOSED simply connected set(s) and remove BOUNDED subsets of the interior.."

Of course that is the same as "the complement of (the union of) the unbounded component(s) of the complement of $S.$"

NOTE: Really I (and you) should say something like "remove a finite number of open subsets each of which has compact closure." Although one could also allow the case of an infinite number of holes, perhaps with disjoint closures. It depends somewhat on the exact situation you wish to model.

LATER: In answer to your actual question: I am not aware of a name for what you want, although that does not at all imply there isn't one. You could call it the simply connected closure and define it as the intersection of all simply connected closed sets containing $S.$ That would definitely work provided that such an intersection is closed and simply connected. It does not immediately give one an intuitive picture of what that thing is.

  • $\begingroup$ Yes, the set I described in that comment should be compact. Besides, your definition is very nice. I guess that I could also define a hole $H_i$ of $S$ as a bounded connected subset of the complement of $S$. Then $H(S)$ is the union of all such sets. Would this be equivalent to the definition I originally posted? $\endgroup$ – jcmonteiro May 14 '16 at 21:19
  • $\begingroup$ I worry that there might be complex topological situations beyond what you (and I) are thinking of. For example let $T$ be the disk of radius $\sqrt{2}$ centered at the origin (to avoid rational points on the boundary.) Let $x_1,\cdots$ be the rational points inside and $H_i$ the open disk of radius $r_i$ around $x_i$ where $r_i$ is the smaller of $4^{-i}$ and half the distance to the boundary. So some of the holes might overlap. Maybe that is not a problem. I'd feel on safer ground if the number of holes was finite or they were convex sets each with positive separation from the others. $\endgroup$ – Aaron Meyerowitz May 14 '16 at 21:50
  • $\begingroup$ Unfortunately, I cannot force the holes to be convex (even though they will be in most cases). Assuming you are constructing the set $S$ by taking the original disk and removing those $H_i$, I don't get the problem of overlapping holes. What I would do is just call those original "removed chunks" $D_i$ and proceed to define the holes as before. Thus, overlapping "chunks" would all fall into the same hole. I would also use the same argument for intersecting $D_i$ (if there are any). That is why I only require a compact connected set. $\endgroup$ – jcmonteiro May 14 '16 at 22:15

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