I was hoping to get a hint for the following problem (4.48 from Mathematical Analysis by T. Apostol which I'm trying to self-study). If $S$ is an open connected set in $\mathbb{R}^n$ and $T$ is a component of $(\mathbb{R}^n-S),$ I'm to show that $(\mathbb{R}^n - T)$ is connected.

I don't have much, just some observations:

  • If I argue, towards contradiction, that $(\mathbb{R}^n - T)$ is disconnected, then $(\mathbb{R}^n - T) = A \cup B$ for $A,B$ disjoint and open in $(\mathbb{R}^n - T)$. Now, we cannot have both $S \cap A$ and $S \cap B$ nonempty, for otherwise $S = (S \cap A) \cup (S \cap B)$ would be disconnected. So I can take $S \subset A$.
  • Since $S$ is open, $(\mathbb{R}^n - S)$ is closed.
  • Here is one other thought that comes to mind: I'm sure there must be something special about $\mathbb{R}^n$ here, but I'm not seeing it. For example, for $n=1$, $S$ is an open interval. Therefore, $(\mathbb{R}-S)$ is either one or two intervals closed at one end and infinite at the other. Then $T$ is all of $(\mathbb{R}-S)$ in the former situation, or exactly one of these halves in the latter. Regardless, this leaves $(\mathbb{R}-T)$ an open interval, which is connected in $\mathbb{R}$. So $n=1$ is reasonably clear, but general $n$ is not obvious to me.
  • $\begingroup$ S is contained in the complement of T. Could the complement of T be open? (not a hint - I don't know). $\endgroup$ Jun 9, 2017 at 0:31
  • $\begingroup$ You need the fact that R^n is connected. $\endgroup$ Jun 9, 2017 at 0:40
  • $\begingroup$ @WilliamElliot Hm, so, in my notation above, if I can prove that X = B $\cup$ T is open in R^n, then it will follow that R^n = A $\cup$ X with A $\cap$ X empty, contradicting connectedness of R^n? Am I on the right track there? $\endgroup$
    – samiam
    Jun 9, 2017 at 1:31
  • $\begingroup$ Hm, probably not? My comment above doesn't reflect the openness of S... $\endgroup$
    – samiam
    Jun 9, 2017 at 1:32
  • $\begingroup$ I just posted an answer. Do you know that if $C$ is any connected set and $C\subseteq D\subseteq\overline C$ then $D$ must also be connected? In particular $\overline C$ is connected, and components are closed sets. It is well-known, but it is used in my answer. If you didn't know it, please see for example this answer math.stackexchange.com/q/441655 ... almost forgot, Welcome to MSE, and please accept and/or vote up my answer, if you find it correct and helpful, or ask for more details! $\endgroup$
    – Mirko
    Jun 9, 2017 at 19:35

2 Answers 2


First prove that if $K$ is any component of $(\mathbb{R}^n-S)$, then $S\cup K$ is connected.

Some details:

It is easily seen that $K$ is closed. It cannot also be open as this would partition $\mathbb{R}^n$. If $p$ is any boundary point of $K$ then $p\in\overline{S}$, for otherwise we could add a small ball around $p$ and obtain a strictly bigger than $K$ connected set missing $S$, a contradiction. It follows that $S\cup\{p\}$ and $K$ are two connected sets with a non-empty intersection, hence their union is connected.

Next represent $(\mathbb{R}^n-T)$ as the union of a family of connected sets that do have a non-empty intersection:

Indeed $(\mathbb{R}^n-T)=\bigcup\{S\cup K: K$ is a component of $(\mathbb{R}^n-S)$ and $K\not=T \}$ .

We do not need to assume that $S$ is open.

If $S$ need not be open, then $K$ (in the notation introduced above) need not be closed in $\mathbb{R}^n$. But, just as before, $K$ cannot be both closed and open, hence it has a non-empty boundary $\mathrm{Bd\,}K$. Pick $p\in\mathrm{Bd\,}K$. If $p\in S$ then $S$ and $K\cup\{p\}$ are two connected sets with a non-empty intersection, hence their union $S\cup K$ is connected. If $p\not\in S$, then $p\in K$ (since $K\cup\{p\}$ is connected and $K$ is a component of $(\mathbb{R}^n-S)$, i.e. a maximal connected subset). As before we must have that $p\in\overline S$, and that $S\cup\{p\}$ and $K$ are two connected sets with a non-empty intersection, hence their union $S\cup K$ is connected.

The above may be generalized as follows.
Suppose that $X$ is a connected topological space and $S$ is a connected subset. Suppose also that $\overline S\cap \overline K\not=\emptyset$ for every component $K$ of $X-S$. Then, if $T$ is any component of $X-S$, we have that $X-T$ is connected.

The condition above that $\overline S\cap \overline K\not=\emptyset$ certainly holds if $X$ is locally connected. Indeed, if $\overline S\cap \overline K=\emptyset$ then $\overline K$ is a connected set missing $S$, hence $K=\overline K$, i.e. $K$ is closed. But $K$ is also open (as a component) in $X-\overline S$ since $X-\overline S$ is locally connected (being open in the locally connected $X$). Then $K$ is both closed and open in $X$, a contradiction.

I do not know if the condition that $X$ is locally connected is necessary. Perhaps $\overline S\cap \overline K\not=\emptyset$ always holds, regardless of whether $X$ is locally connected or not? I posted this as a separate question .

Edit. My separate question was answered (with a link to the answer of another older question). There is indeed a space $X$ (that is not locally connected at just two points), a connected subset $S$ (which in that example is a singleton $\{(0,1)\}$, but could be made open by taking a small neighborhood), and a component $K$ of $X\setminus S$ (namely $K=\{(0,0)\}$ again a singleton) such that $\overline S\cap \overline K=\emptyset$.

Question (which I will not post for now as it is getting too late, please feel free to post separately if you wish). Assume that $X$ is a connected topological space, $S$ is a connected subset, and $T$ is a component of $X\setminus S$. Is $X\setminus T$ connected? My particular proof here does not work in general, but perhaps the answer is nevertheless yes, with another proof?

  • $\begingroup$ Here is one attempt at proving the 1st claim (I'll try to wait a little bit before I look at the spoiler): Suppose that $S \cup K$ is disconnected; then write $S \cup K = A \cup B$ for $A, B$ disjoint, open in $S \cup K$. Then without loss of generality we must have $S = A$ and $K = B$ since $S, K$ are disjoint, connected sets. But then $K$ is open and closed, contradicting connectedness of $\mathbb{R}^n$. $\endgroup$
    – samiam
    Jun 9, 2017 at 20:16
  • $\begingroup$ Hm, actually I think my attempt is incorrect, for the following reason: I'm not being careful about what 'open' means here. In particular, $K = B$ does not imply that $K$ is open and closed in $\mathbb{R}^n$. This is because $B$ is open...but open in $S \cup K$--not necessarily in $\mathbb{R}^n$. Therefore I'm not allowed to say that $K$ is clopen in $\mathbb{R}^n$, right? $\endgroup$
    – samiam
    Jun 9, 2017 at 20:19
  • $\begingroup$ Alright, I got eager and looked--wish I came up with this proof, it's very nice and I understand it. I'll accept this one. Thanks. $\endgroup$
    – samiam
    Jun 9, 2017 at 20:30
  • 1
    $\begingroup$ The intuition seem to be that $S$ union $K$ is like filling in pot holes. Every fix is an 'improvement'. Give @mirko the contracting job! $\endgroup$ Jun 9, 2017 at 23:45
  • 1
    $\begingroup$ @samiam actually maybe I didn't know how to deal with the case that $T$ might be open ... don't remember, but somehow at one point all fit in place. Just wanted to say that what I appreciate in your writing, is that you are able to catch your mistakes and know when a proposed proof is not quite a proof, not many people possess this ability, and it, in the long run, I think, develops the ability to come up with new proofs (as one keeps searching for a correct and complete solution and has to explore various options and possibilities). Best wishes and good luck :) $\endgroup$
    – Mirko
    Jun 9, 2017 at 23:45

Stuff like your problem was discussed some years ago at Ask
a Topologist, topology forum where it resides in the archives.

Since S is open and R^n both connected and locally connected, there may be some short cuts to these results from my notes,
each dependend upon the previous. The last generalises your
problem. Proofs are not long but in all it's complicated.

A,B separated when A cap cl B and B cap cl A are empty.
S is a topological space with K subset U.

1 separated A,B, closed K, S\K = A cup B implies A cup K closed
2 connected S,K, separated A,B, S\K = A cup B implies A cup K connected
3 connected S,K, A clopen within S\K implies A cup K connected
4 connected S,K, C connected component S\K implies S\C connected


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