Is the submanifold compact? Let $M$ be the following subsets of $\mathbb R^4$:$$M= \{(x,y,z,w), 2x^2+2=z^2+w^2, 3x^2+y^2=z^2+w^2 \}$$
we know $M$ is a submanifold of $\mathbb R^4$, is $M$ compact?
 A: Yes, because $M$ is closed and bounded. 
Closedness: 
$M$ is given implicitly on the form $F=0$ with $F:\mathbb{R}^4\rightarrow\mathbb{R}^2$, $F$ continuous. Continuiuty guarantees closedness. To see this, recall that a set is closed if and only if it contains all its limit points. Let $u_n = (x_n,y_n,z_n,w_n) \in \mathbb{R}^4$ be a sequence on $M$, i.e., $F(u_n) = 0$ for all $n$. If $u_n$ converges to some $u\in\mathbb{R}^4$, it follows by continuity of $F$ that $F(u)=0$. Thus, the limit point $u\in M$ as well.
Boundedness: The defining equations $F$ can be combined to give
$$ 3x^2+y^2 = x^2 + w^2 = 2x^2 + 2$$
and hehce
$$ x^2 + y^2 = 2. $$
Thus the projection of $M$ on the $xy$ plane is a circle of radius $\sqrt{2}$. Hence, $x^2$ and $y^2$ are smaller than 2 for every point $(x,y,z,w)\in M$. Similarly,
$$ z^2 + w^2 = 2 + 2x^2 \leq 6, $$
so the projection of $M$ on the $zw$ plane is contained in a circle of radius $\sqrt{6}$. Thus, both $z^2$ and $w^2$ are smaller than 6.
It follows that for all $(x,y,z,w)\in M$, $x^2+y^2+z^2+w^2 \leq 8$, and $M$ is bounded.
