Let $R > r > 0$ and $A = \{ (x,y,z): r^2 \leq x^2+y^2+z^2 \leq R^2\}$ show that $A$ is path-connected

Question

Let $R > r > 0$ and $A = \{(x,y,z): r^2 \leq x^2+y^2+z^2 \leq R^2\}$ show that $A$ is a path-connected. A path from a point $x$ to a point $y$ in a topological space $X$ is a continuous function $ƒ$ from the unit interval $[0,1]$ to $X$ with $f(0) = x $ and $ f(1) = y$.

I can resolved that in $\Bbb R^2$ but i can´t find the path in $\Bbb R^3$

Answer 1

Let $S = \{ x \in \mathbb{R}^3 | \|x\|=R \}$.

Given any point $x \in A$ the path $p(t) = (t {R \over \|x\|} +(1-t))x $ joins $x$ to the point ${R \over \|x\|}x \in S$.

So, we can presume that we have two points $x,y \in S$. If the line joining $x,y$ passes through the origin, pick some other point $z \in S$ that is orthogonal to $x$. It is straightforward to show that the segments $[x,z]$ and $[z,y]$ do not pass through the origin.

So, we can presume that the segment $[x,y]$ does not pass through the origin.

Then the path $p(t) = {R \over \|t y +(1-t)x\|} (t y +(1-t)x)$ lies in $S$ and connects $x,y$.