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I have to show that $\mathbb{R}^{p}\backslash \{0 \}$ is connected for $p \geq 2$. Is it possible to show this using the property debated on in this article: Union of connected subsets is connected if intersection is nonempty?

ajotatxe
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simp
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  • You mean $p\color{red}\ge2$ – Vim Jun 16 '17 at 23:18
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    Well it surely can be done by observing that it is path-connected and hence connected – asdf Jun 16 '17 at 23:19
  • @coconut Their intersection is empty: namely $\bigcap_i U_{i,+}\cap U_{i,-}=\underbrace{\emptyset\cap\cdots\cap\emptyset}_{p\text{ times}}=\emptyset$. –  Jun 17 '17 at 00:48

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$\Bbb R^p\setminus\{0\}$ is the union of the $2p$ open halfspaces $x_1>0$, $x_1<0$, $x_2>0$, $x_2<0$, $\ldots$, $x_p>0$, $x_p<0$.

Halfspaces are convex and hence connected.

EDIT: Certainly, the intersection of all the halfspaces is empty. But they still make the job:

Indeed: let $H_j^+=\{x_j>0\}$ and $H_j^-=\{x_j<0\}$, for $j\in\{1,\ldots,p\}$. Define: $$A_1=H_1^+,\; A_{j+1}=A_j\cup H_{j+1}^+\text{, for }2\le j\le p-1$$ $$A_{p+1}=A_p\cup H_1^-,\; A_{p+j+1}=A_{p+j}\cup H_{p+j+1}^-\text{, for }2\le j\le p-1$$

Then each $A_j$ is connected because is the non-disjoint union of two connected sets. And $A_{2p}=\Bbb R^p\setminus\{0\}$.

Note that this reasoning does not work (as it should be) if $p=1$ because $A_1\cap A_2=\emptyset$.

ajotatxe
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As observed, you can cover $\Bbb R^p\setminus\{0\}$ with half-spaces in the natural way.

However, said covering does not satisfy the hypothesis of the theorem you've linked: there are in fact $p$ couples of disjoint subsets!

To proceed with that idea, I suggest a slight variation of the theorem you've linked (which you can prove as an exercise):

Let $X$ be a topological space and let $\mathfrak U$ be a covering of $X$ in non-empty connected subspaces. If for any $A,B\in \mathfrak U$ such that $A\cap B=\emptyset$ there is $C\in \mathfrak U$ such that $C\cap A\ne\emptyset$ and $C\cap B\ne\emptyset$, then $X$ is connected.