# Proving that $\Bbb R^n-T$ is connected

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?

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 $$T_0$$.