# Math Help - connectivity question in R^{n}

1. ## connectivity question in R^{n}

This is an exercise (4.48) from Apostol's book on analysis. S is an open connected set in $R^{n}$. Let T be a component of R^{n} - S. Prove that $R^{n} - T$ is connected.

I take it that the appropriate formulation of 'connected' for subsets of a topological space M is that $X \subset M$ is connected iff for any partition $X \subset A \cup B$, where A and B are nonempty disjoint sets open in M, either $X \subset A$ or $X \subset B$, and I know that a component of X is a maximally connected subset of X ie a subset $Y \subset X$ such that for all $Y \subset Z \subset X$, Z is connected iff $Z = Y$.

I can of course write down the algebraic formulation of the hypotheses of the problem, but don't see where to go with them. And I am sadly bereft of any geometric intuition to guide me.

Supplementary query: how do I get the LaTex symbols to work in the above? I am completely new to all this.

2. Originally Posted by leopold
This is an exercise (4.48) from Apostol's book on analysis. S is an open connected set in $R^{n}$. Let T be a component of R^{n} - S. Prove that $R^{n} - T$ is connected.

I take it that the appropriate formulation of 'connected' for subsets of a topological space M is that $X \subset M$ is connected iff for any partition $X \subset A \cup B$, where A and B are nonempty disjoint sets open in M, either $X \subset A$ or $X \subset B$, and I know that a component of X is a maximally connected subset of X ie a subset $Y \subset X$ such that for all $Y \subset Z \subset X$, Z is connected iff $Z = Y$.

I can of course write down the algebraic formulation of the hypotheses of the problem, but don't see where to go with them. And I am sadly bereft of any geometric intuition to guide me.

Supplementary query: how do I get the LaTex symbols to work in the above? I am completely new to all this.
If $S=\mathbb{Re}^n$ or empty set, it is trivial to show that $\mathbb{Re}^n - T$ is connected. Assume not.
Let X be $\mathbb{Re}^n - S$ and let Y be $\mathbb{Re}^n - T$; let $T_k$ be a component for X where k=1,2, ... , m and m is the number of components in X. We see that $\overline{S} \cap T_k \neq \emptyset$ for every k=1,2, ... , m. Without loss of generality, $Y=\mathbb{Re}^n - T = \mathbb{Re}^n - T_1 = T_2 \cup T_3 \cup \cdots \cup S$. Suppose to the contrary that $Y=\mathbb{Re}^n - T_1$ is not connected. Then there exists a clopen component of Y except the empty set and Y itself. If S is such a component, then $Y=S \cup (Y-S)$ where S and Y-S are disjoint open sets such that $\overline{S} \cap (Y-S) = \emptyset$, contradicting that $\overline{S} \cap T_k \neq \emptyset$ for every k=1,2, ... , m, where $T_k \subseteq (Y-S)$. If $T_j$ for $1 \leq j \leq m$ is such a component, then $Y=T_j \cup (Y-T_j)$ where $T_j$ and $Y-T_j$ are disjoint open sets such that $T_j \cap \overline{Y-T_j} = \emptyset$, contradicting that $T_k \cap \overline{S} \neq \emptyset$ for every k=1,2, ... , m, where $\overline{S} \subseteq \overline{Y-T_j}$.
Thus $Y=\mathbb{Re}^n - T$ is connected.

3. aliceinwonderland

1. Why is for all k? I assume you are using some well-known result about components here (I note for example that you haven't explicitly appealed to the maximality of T among connected subsets of ), but it's not well-known to me and despite some effort I have not succeeded in reconstructing it.

2. (May well be related to (1.)) In the parts (two alternative forks) of the argument in which you are deriving a contradiction, you say (in one place) ; and (in the other) . Again, it's not clear to me why these hold. (I thought at first that eg because (as you said) Y - S is open (and so S is closed). But then I realised that S is only closed in the induced subspace topology on Y, not in )

Also, may I ask how you got to this proof? I would be particularly interested if it was inspired by some kind of intuition, since I have no intuition at all regarding this stuff and would like to develop some. All I have been doing in tackling this problem is manipulating symbolic expressions in the (vain) hope that something might fall out.

Thanks again.

4. Originally Posted by leopold
aliceinwonderland

1. Why is for all k? I assume you are using some well-known result about components here (I note for example that you haven't explicitly appealed to the maximality of T among connected subsets of ), but it's not well-known to me and despite some effort I have not succeeded in reconstructing it.
There are some theorems that are directly or indirectly used in the proof.

Theorem 1. Let A be a subset of a topological space W. Then A is open and closed if and only if $\text{bdy A} = \emptyset$.

Since S is a proper open subset (excluding an empty set) of $\mathbb{Re}^n$, S cannot be both open and closed because $\mathbb{Re}^n - S$ cannot be open. Thus S cannot have an empty boundary by theorem 1. By definition of boundary, $\overline{S}$ has an intersection with $\mathbb{Re}^n - S$.

For instance, if we consider $\mathbb{Re}^2$ with $S=\{(0,1) \times \mathbb{Re}\} \cup \{\mathbb{Re} \times (0,1)\}$, then $\mathbb{Re}^2 - S$ has four components, where each component has an intersection with $\overline{S}$.
From this observation, if we remove a component $T_k$ from $\mathbb{Re}^n$ and consider the subspace topology, $\overline{S}$ still has intersections with $T_j$ with $j \neq k$. Based on this observation, I had established that $\overline{S} \cap T_k \neq \emptyset$ for every k=1,2, ... , m.
I relied upon this observation rather than using well-known theorems about components. Can you find any counterexample here?

2. (May well be related to (1.)) In the parts (two alternative forks) of the argument in which you are deriving a contradiction, you say (in one place) ; and (in the other) . Again, it's not clear to me why these hold. (I thought at first that eg because (as you said) Y - S is open (and so S is closed). But then I realised that S is only closed in the induced subspace topology on Y, not in )
Note that for a closed set or clopen set A in Y, $\overline{A}=A$.
If we assume Y is not connected for the sake of contradiction, then Y has a proper component A having an empty boundary in the topological space Y. By using the previous established result that $\overline{S} \cap T_k \neq \emptyset$ for every k=1,2, ... , m in $\mathbb{Re}^n$ (those intersections do not vanish in the subspace of $\mathbb{Re}^n$ with $T_k$ removed), we can draw a necessary contradiction and conclude that Y is connected.

Also, may I ask how you got to this proof? I would be particularly interested if it was inspired by some kind of intuition, since I have no intuition at all regarding this stuff and would like to develop some. All I have been doing in tackling this problem is manipulating symbolic expressions in the (vain) hope that something might fall out.

Thanks again.
To show that a topological space is connected we often assume that a topological space is not connected and derive a contradiction. I used boundary properties to derive a necessary contradiction.
Hope you find a better solution than mine .

5. Thanks alice. But although I see that has a nonempty intersection with , I am afraid I still do not understand your reasoning that it must have a non-empty intersection with any component (ie any maximal connected subset) of .

(Your question whether I can find any counterexample worries me somewhat; are you claiming to have a proof of this, or not?)

Apologies if I am coming over as, so to speak, everywhere dense.

6. Originally Posted by leopold
Thanks alice. But although I see that has a nonempty intersection with , I am afraid I still do not understand your reasoning that it must have a non-empty intersection with any component (ie any maximal connected subset) of .

(Your question whether I can find any counterexample worries me somewhat; are you claiming to have a proof of this, or not?)

Apologies if I am coming over as, so to speak, everywhere dense.
Hint. $\mathbb{Re}^n$ is a connected set and each component of $\mathbb{Re}^n$ is a closed subset of $\mathbb{Re}^n$ (link). Apply induction by increasing the number of components and observe their intersections with the closure of their complements in $\mathbb{Re}^n$.

7. First, I don't understand what the inductive hypothesis might be.

Second, I don't understand how induction can help here anyway. After all, there might be an infinite number of components of . (Or did you mean transfinite induction? But I find it hard to believe that can be necessary here)

8. Originally Posted by leopold
First, I don't understand what the inductive hypothesis might be.

Second, I don't understand how induction can help here anyway. After all, there might be an infinite number of components of . (Or did you mean transfinite induction? But I find it hard to believe that can be necessary here)
Lemma 1. $\mathbb{Re}^n$ is a connected set.
Lemma 2. The connected components are always closed.
Lemma 3. Let X be a topological space. Then,
X is connected <-> The only subsets of X which are clopen are X and empty set <-> The only subsets of X with empty boundary are X and the empty set.

Lemma 2 can be proven by using the maximal property of a connected compoent (if A is a connected set, then $\bar{A}$ is also a connected set). Lemma 1,2, and 3 are already discussed in my previous replies and you can find the above lemmas here.

Claim. Let S be a proper nonempty proper open subset of $\mathbb{Re}^n$. Let $X=\mathbb{Re}^n - S$. Let $T_k$ be a component of X, where $k \in I$ for an index set I. Then $\bar{S} \cap T_k \neq \emptyset$ for all $k \in I$.

Suppose the claim is false. Then there exists $j \in I$ such that $\bar{S} \cap T_j = \emptyset$. Then, the union of $\bar{S} \cup \bigcup_{i \in (I-\{j\})}T_i$ and $T_j$ is $\mathbb{Re}^n$ and the intersection of them is empty by hypothesis and by the fact that each distinct component is disjoint in the original space. The former is a closed set by lemma 2 and by the fact that the arbitrary union of closed sets is closed. It follows that $T_j$ is a clopen set in $\mathbb{Re}^n$, contradicting lemma 1 and lemma 3. Thus the claim is true.

I used an arbitrary index set I this time to take an uncountable index set into account. I think the transfinite induction should also work in this case as well. Here is the idea:
Let an index set I be well-ordered. If $\alpha$ is a limit ordinal, let $T'_\alpha = \bigcup_{\beta < \alpha}T'_\beta$ such that the $\bar{S} \cap T'_\beta \neq \emptyset$ for all $\beta$ where $T_i' \subset T_j'$ if $i < j$ in I and $T'_k$ for $k \in I$ is an arbitrary union of the connected comonents of $\mathbb{Re}^n - S$. Then $T'_\alpha$ holds the above claim, otherwise, contradicting lemma 1,2, and 3.

9. Sorry for the delay in getting back to this. (Festive obligations, dontchaknow...)

I now see that for all in the case where the index set I is finite. But in the case where it is infinite (not even uncountable) I don't see how to get this result. In the proof of your claim you adverted to "the fact that the arbitrary union of closed sets is closed", but this is wrong, isn't it? The arbitrary union of open sets is open, and the arbitrary intersection of closed sets is closed, but not {what you said}. I wondered for a while whether your discussion of the limit ordinal case might help here, but I didn't understand it well enough to see how!

10. Originally Posted by leopold
Sorry for the delay in getting back to this. (Festive obligations, dontchaknow...)

I now see that for all in the case where the index set I is finite. But in the case where it is infinite (not even uncountable) I don't see how to get this result. In the proof of your claim you adverted to "the fact that the arbitrary union of closed sets is closed", but this is wrong, isn't it? The arbitrary union of open sets is open, and the arbitrary intersection of closed sets is closed, but not {what you said}. I wondered for a while whether your discussion of the limit ordinal case might help here, but I didn't understand it well enough to see how!
That is my stupid mistake. Thanks for pointing that out.

Anyhow, I don't find any evidence that the above claim of mine is false.
$\bar{S} \cup \bigcup_{i \in I-\{j\}}T_i$ is a still closed set (\bar{S} has an intersection with each T_i (i not equal j) by hypothesis). Actually, it is a single component by maximality and every component is a closed set as shown in the above lemma. In the above proof for the claim, note that

$\neg \forall{x} P(x) \leftrightarrow \exists{x} \neg P(x)$,

This holds whether the universe of the quantified variables is countable or uncountable.