# Components and connectedness

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.
• S is contained in the complement of T. Could the complement of T be open? (not a hint - I don't know). Jun 9, 2017 at 0:31
• You need the fact that R^n is connected. Jun 9, 2017 at 0:40
• @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? Jun 9, 2017 at 1:31
• Hm, probably not? My comment above doesn't reflect the openness of S... Jun 9, 2017 at 1:32
• 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! Jun 9, 2017 at 19:35

Hint.
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 $K\not=T \}$ .

Edit.
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?

• 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 $\mathbb{R}^n$. Jun 9, 2017 at 20:16
• 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? Jun 9, 2017 at 20:19
• 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. Jun 9, 2017 at 20:30
• 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! Jun 9, 2017 at 23:45
• @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 :) Jun 9, 2017 at 23:45