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Theorem: $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^{n} \forall n>1$

To prove this, all that is required is to show that $\mathbb{R}\setminus {0}$ is separated and $\mathbb{R}^{n}\setminus {y}$ is connected. However, I'm facing a bit of an issue as to how it can be shown that $\mathbb{R}^{n}\setminus {y}$ is path connected.

Any help is appreciated.

Thanks in advance.

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Without loss of generality, let $y$ be $0$ in $\mathbb R ^n$; now you can connect two points with a convex combination if the segment that connects them does not pass through the origin; if it passes through the origin, than consider a plane that contains the two points and the origin, now you can consider the triangle that connects the two points and not passes through the origin, so it's done.

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$\Bbb R\setminus\{0\}$ is disconnected whereas $\Bbb R^n\setminus\{0\}$ is not.

$\Bbb R^n\setminus\{0\}$ is in fact path connected.

To see this take any two points $a,b\in \Bbb R^n\setminus\{0\}$. We can find uncountable number of straight lines passing through $a$.

Now it is quite obvious to find a straight line joining through $a$ and $b$ without passing through $0$.

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  • $\begingroup$ I don't find it quite as obvious as you do $\endgroup$ – cronos2 Nov 12 '16 at 16:13
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$\textbf{Fact}$: The product of two path connected spaces is path connected: $\mathbb{R}^n \setminus \{y\} \cong S^{n-1} \times \mathbb{R}$

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