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Let $S$ be an open connected set in $\Bbb R^n$, let $T$ be a component of $\Bbb R^n-S$. Prove that $\Bbb R^n-T$ is connected

Proof: Suppose, $\Bbb R^n-T$ is not connected. Then, we may write it as $R^n-T=A\cup B$, where $A,B$ are non empty disjoint open connected sets. Now,we have, $\Bbb R^n=B\cup A\cup T$

$R^n$ being connected we must have, $\bar{A}\cap T$ or $\bar{B}\cap T$ is non empty. Say, $\bar{A}\cap T$ is non empty.

So, $A\cup T$ is a connected set.

Then, $A$ can't be a subset of $\Bbb R^n-S$ as $T$ is a component of $\Bbb R^n-S.$ So, $S\subset A$.

Therefore, $\bar{B}\cap T=\phi$ which means $A\cup T$ and $B$ are separated but $\Bbb R^n$ is a connected set.

Is it correct?

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If $T\cap\text{cl}(S)=\emptyset$ we could separate the disjoint closed connected sets $\text{cl}(S),T$ with disjoint open connected sets $U_T,V_T\subset\Bbb R^n$, i.e. $\text{cl}(S)\subset U_T,\;T\subset V_T\subset\text{cl}(V_T)$, but then $\text{cl}(V_T)$ would be closed and connected in $\Bbb R^n-S$ violating the maximality of the component $T$. Therefore, $T\cup S$ is always connected and consequently so is $$\bigcup_{T\neq T_0}(T\cup S)$$ where $T$ takes all $\Bbb R^n-S$ component values except some fixed $T_0$.

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