Show that $M = \{(x,y) \in \mathbb{R}^p \times \mathbb{R}^q : \lvert x \rvert = \lvert y \rvert \neq 0 \}$ is connected for $p,q \geq 2$

How can I show that $M = \{(x,y) \in \mathbb{R}^p \times \mathbb{R}^q : \lvert x \rvert = \lvert y \rvert \neq 0 \}$ is connected?

I can find a path $\gamma(t) = (tx, ty)$ which lays on $M$ for $t$ in $[1,\frac{1}{\| x \|}]$, then we have that

$\gamma(1) = (x,y)$ and $\gamma \left(\frac{1}{\| x \|} \right) = \left(\frac{x}{\| x \|}, \frac{y}{\| y \|}\right)$.

How do I have to conclude from here? Thanks!

Solution 1:

(It seems to be implied by your own work that you know that path-connected spaces are connected.)

What you've done shows that there is a path in $M$ from an arbitrary $(x, y) \in M$ to the pair $\left(\frac{x}{\lVert x \rVert}, \frac{y}{\lVert y \rVert}\right) \in M$, which in particular are unit vectors. So, it's enough to now show that any two pairs $(x, y)$ and $(x', y')$ in $M$ which both have $\lVert x \rVert = \lVert x' \rVert = 1$ are connected by a path in $M$.

That's not so hard: we know that the unit sphere $S^{p - 1}$ in $\mathbb{R}^p$ is path-connected for $p \geq 2$, so there exists a path $\gamma : [0, 1] \to S^{p-1}$ from $x$ to $x'$, and likewise a path $\gamma' : [0, 1] \to S^{q-1}$ from $y$ to $y'$. Thus the map $t \mapsto (\gamma(t), \gamma'(t))$ is a path in $M$ from $(x, y)$ to $(x', y')$, as desired.