Let $K$ be a compact subset of an arbitrary open set $\Omega\subset \mathbb{R}^n.$ It is said that a connected component $W$ of $\Omega\setminus K$ is $\Omega$-bounded if $\overline{W}$ is a compact subset of $\Omega$.

Heuristically speaking, the $\Omega$-bounded components of $\Omega\setminus K$ are the "holes" of $K$ that do not contain "holes" of $\Omega$. Now, I want to prove this, but I am lost. Topology is not my usual work field.

Specifically, I want to prove that if $W$ is an $\Omega$-bounded component of $\Omega\setminus K$ and $H$ is the component of $\mathbb{R}^n\setminus K$ that contains $W,$ then $H\subset \Omega.$ Additionally, is it possible to prove that $H=W$?

  • $\begingroup$ Alex M. But in your example $\overline{W}$ is not contained in $\Omega$, which is the starting point: $\overline{W}$ must be a compact set included in $\Omega$ for $W$ to be called $\Omega$-bounded. $\endgroup$
    – user120667
    Commented Feb 11, 2018 at 17:15
  • $\begingroup$ I realized this only after having posted my comment. Anyway, the tag "approximation-theory" has nothing to do with the question. $\endgroup$
    – Alex M.
    Commented Feb 11, 2018 at 17:17
  • $\begingroup$ Yes, you're right, even though this condition does appear in connection with approximation theory. If your compact set $K$ does not have $\Omega$-bounded componentes, then you can approximate functions defined on $K$ by functions defined on $\Omega$. The functions involved having certain types of regularity. $\endgroup$
    – user120667
    Commented Feb 11, 2018 at 17:22
  • $\begingroup$ What do you mean by "$\overline{W}$ is a compact set of $\Omega$"? In particular, do you mean the closure in $\mathbb{R}^n$ or the closure in $\Omega$? And secondly, what is "a set of $\Omega$"? $\endgroup$ Commented Feb 11, 2018 at 17:38
  • $\begingroup$ The closure is taken in $\mathbb{R}^n$ and $\overline{W}\subset\Omega$ $\endgroup$
    – user120667
    Commented Feb 11, 2018 at 17:41

1 Answer 1


Yes, $W=H$, so in particular $H \subset \Omega$.

(All uses of closure, boundary, open, closed, complement, etc, are relative to $\mathbb{R}^n$ unless otherwise stated.)

First, note that for any open subset $U \subset \mathbb{R}^n$, a connected subset of $U$ is a component of $U$ iff it is open and relatively closed in $U$. This follows from the local connectedness of $\mathbb{R}^n$.

Claim. $\partial W \subset K$.

Proof. Let $x \in \partial W$. Clearly $x \notin W$ since $W$ is open. Since $W$ is $\Omega$-bounded, we have $x \in \Omega$. If $x \notin K$, then $x$ is in some other component $W'$ of $\Omega \setminus K$. But $W'$ is open and hence $W'$ is not disjoint from $W$, which is absurd. $\Box$

Now since $W$ is open, the claim implies that $W = \overline{W} \setminus \partial W = \overline{W} \cap K^c$. So $W$ is relatively closed in $K^c$. Being connected and relatively clopen means that $W$ is a connected component of $K^c$, hence it equals the component $H$ which contains it.

Note we didn't use the compactness of $K$ (closed would suffice) nor of $\overline{W}$.

  • 1
    $\begingroup$ Just a slight distillation: let $X$ be a locally connected topological space, let $U\subset X$ be open, let $K\subset U$ be closed, let $W$ be a component of $U-K$ such that $\overline{W}\subset U$. Then $W$ is a component of $X-K$. Indeed, since $X$ is locally connected the components of open sets are open, and so every component of $U-K$ is open. Then the union $V$ of all the components other than $W$ is also open, and so $\overline{W}\cap V=\varnothing$. Hence, $\partial W\subset U-(W\cup V)=K$. Now $X-K\subset W\cup (X-\overline{W})$, and so $W$ is a component of $X-K$. $\endgroup$
    – erz
    Commented Feb 11, 2018 at 23:28

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