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$\Bbb R^n-S=T\cup (\bigcup A)$, $T$ and these $A$ are components in $\Bbb R^n-S$.

$\Bbb R^n-T=S\cup(\bigcup A)$,

$\DeclareMathOperator{\cl}{cl}$ since $\partial S \subseteq \Bbb R^n-S,\partial S\subseteq (\bigcup A)$

I consider $S\cup \partial S \cup (\bigcup A)=\bigcup (\cl S \cup A)$

If $S$ is not an empty set, and if for every $A$, $\cl S \cup A$ is connected, in that way, $\Bbb R^n-T$ is connected.

Closure of $S$ is connected, every $A$ is connected,if $\cl S\cap A$ is not empty set, in that way, $\cl S\cup A $ is connected.

But I can’t prove $\cl S\cap A $ is not the empty set. Could you give me some hints or other methods different from mine to solve this problem? Thank you in advance

The definition of component

every metric space $S$ can be expressed in a unique way as a union of connected "pieces" called components.

Sorry, I found a mistake of my method, $\partial A\subseteq \bigcup A\cup T$, and my method is wrong now.

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  • $\begingroup$ What is the context of your question? Is it an exercise in general topology? $\endgroup$
    – Paul Frost
    Commented Jul 23, 2019 at 8:23
  • $\begingroup$ It’s an exercise in Tom Apostol mathematical analysis page100,exercise 4.48 $\endgroup$
    – user682113
    Commented Jul 23, 2019 at 8:25
  • $\begingroup$ @PaulFrost,what do you think this problem?Could you give me some hints? $\endgroup$
    – user682113
    Commented Jul 23, 2019 at 8:35
  • $\begingroup$ Can you show that $T$ is closed and that $S\cup T$ is open? Then if $U,V$ cover $\Bbb R^n-T$, one of them, $U$ say, contains $S$. Replace $U$ with $U\cup T$ and $V$ with $V-T$ and use that $\Bbb R^n$ is connected $\endgroup$ Commented Jul 23, 2019 at 9:14
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    $\begingroup$ $T$ is always closed in $\mathbb R^n$. See my answer. $\endgroup$
    – Paul Frost
    Commented Jul 23, 2019 at 9:39

1 Answer 1

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$\DeclareMathOperator{\cl}{cl}$

Let $A = \mathbb R^n \setminus S$ which is closed in $\mathbb R^n$. All components $C_\iota$ of $A$ are closed in $A$, hence closed in $\mathbb R^n$.

The cases $S = \emptyset, \mathbb R^n$ are trivial, so let us assume $\emptyset \ne S \ne \mathbb R^n$. Then we have $S \subsetneqq \cl S$ because otherwise $S$ would be open and closed in $\mathbb R^n$, i.e. $S = \emptyset$ or $S = \mathbb R^n$ since $\mathbb R^n$ is connected. Thus $\cl S \cap A \ne \emptyset$.

We have $\cl S \cap C_\iota \ne \emptyset$ for all components of $A$. Assume that $\cl S \cap C_\iota = \emptyset$ for some $\iota$. Then $C_\iota$ must be a component of $B = \mathbb R^n \setminus \cl S$. To see this, let $D$ be the component of $B$ containing $C_\iota$. Then $D$ is a connected subset of $A \supset B$ and therefore contained in a component $C_{\iota'}$ of $A$. We have $C_\iota \subset D \subset C_{\iota'}$ which implies $\iota' = \iota$. But now $B$ is open in $\mathbb R^n$, hence locally connected, and this implies that $C_{\iota}$ is open in $B$, hence open in $\mathbb R^n$. Concerning local connectedness see my answer to About local linear connect. We have seen that $C_\iota$ is open and closed in $\mathbb R^n$, hence either $C_\iota = \emptyset$ or $C_\iota =\mathbb R^n$. Both is impossible and our assumption must be false.

We have $\mathbb R^n \setminus T = S \cup \bigcup_{\iota \in I} C_\iota$, where $I$ contains all $\iota$ such that $C_\iota \ne T$. For each $\iota \in I$ pick $x_\iota \in \cl S$ such that $x_\iota \in C_\iota$. Then $S' = S \cup \{x_\iota \mid \iota \in I \}$ is connected because $S \subset S' \subset \cl S$. But now $$\mathbb R^n \setminus T = S' \cup \bigcup_{\iota \in I} C_\iota$$ This show that $\mathbb R^n \setminus T$ is connected because all $S' \cap C_\iota \ne \emptyset$.

Remark.

In this proof one can replace $\mathbb R^n$ by any space $X$ which is connected and locally connected.

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