A double cone in $\mathbb{R}^3$ is path connected.

Question

I want to show that the following subset of $\mathbb{R}^3$ is path connected. Define a double cone as below

$$M=\{(x,y,z)\in\mathbb{R}^3\mid x^2+y^2=z^2\}.$$

The only thing I know is the definition of path-connected. $M$ is path-connected if and only if for every two points $p_1,\,p_2\in M$ there is a continuous map $f:[t_1,t_2]\to M$ such that $f(t_1)=p_1$ and $f(t_2)=p_2$. I can intuitively say that $M$ is indeed path-connected by looking at $M$ in $\mathbb{R}^3$ which looks like below, but I don't how to right down a proof for it! Any help or hint is appreciated.

$\qquad\qquad\qquad\quad$

Answer 1

Given two points $p_1,p_2\in M$, we can take as a path the line segment from $p_1$ to the origin, followed by the line segment from the origin to $p_2$. Explicitly, we have the path $f:[0,1]\to M$ given by

$$f(t)=\begin{cases} (1-2t)p_1 & 0\leq t\leq 1/2 \\ -(1-2t)p_2 & 1/2<t\leq 1 \end{cases}$$

Note that because the defining equation $x^2+y^2=z^2$ of $M$ is homogeneous, any multiples of $p_1$ and $p_2$ are also in $M$, so the path is contained in $M$. The path is also clearly continuous, since it is piecewise linear and the two pieces meet at the origin.

Here is a plot showing this path for two random points on the surface

Answer 2

Here, I try to elaborate on the comment of Sangchul Lee. We shall use the following facts.

1. $\mathbb{R}$ and $[a,b]$ are path-connected.
2. Cartesian product of path-connected sets is path-connected.
3. Continuous image of a path-connected set is path-connected. Indeed, if two sets are homeomorphic and one of them is path-connected then the other one is also path-connected. Path-connectedness is a topological property.
4. A function $f:\mathbb{R}^m\to\mathbb{R}^n$ takes a $x=(x_1,\dots,x_m)\in\mathbb{R}^m$ and gives a $y=(y_1,\dots,y_n)\in\mathbb{R}^n$. Indeed, $y=f(x)$ or equivalently $(y_1,\dots,y_n)=f\big((x_1,\dots,x_m)\big)$. This suggest that we can have $y_i=f_i((x_1,\dots,x_m))$ for $i=1,\dots,n$ and $f=(f_1,\dots,f_n)$. The function $f:\mathbb{R}^m\to\mathbb{R}^n$ is continuous if and only if the functions $f_i:\mathbb{R}^m\to\mathbb{R}$ are continuous.

Now, one can easily verify that the function $f:\mathbb{R}\times[0,2\pi]\to\mathbb{R}^3$ defined below is continuous and its range is $M$. Since its domain is path-connected then its range is also path connected.

$$f(z,\theta)=(z\cos\theta,z\sin\theta,z)$$