Is $R^4\setminus S^3$, with the subspace topology on $R^4$, connected?

Question

$S^3=\{(x_1, x_2, x_3, x_4)\in\mathbb{R}^4\mid x_1^2+x_2^2+x_3^2+x_4^2=1\}$

Is $R^4\setminus S^3$, with the subspace topology on $R^4$, connected?

Answer 1

You have the idea in the comment you posted. The two sets $\{x \in \mathbb{R}^4 \::\: |x| < 1\}$ and $\{x \in \mathbb{R}^4 \::\: |x| > 1\}$ are clearly disjoint and their union is $\mathbb{R}^4 \setminus S^3$. If you prove these facts, you've shown that $\mathbb{R}^4 \setminus S^3$ is indeed disconnected.

Answer 2

You could try this: the set $[0,1) \cup (1,\infty)$ in $\mathbb R$ is not connected, but it is a continuous image of your set.

Answer 3

A picture is worth 996 words:

As for the other four words you will be needing:

The Reverse Triangle Inequality

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So the outside of the unit sphere is open and the open ball about the origin is open.

Voilà - you gotta disjoint union of two open sets!