Showing $x_1 + x_2 + ... + x_n = a$ is compact if $x_i \geq 0$ I was wondering how we can show something like $x + y + z = 6$ is compact. Specifically, I would like to extend it to any number of variables. Cleary it's bounded as $0 \leq x_i \leq a$, bug I don't know how to show it contains all of it's boundary points in arbitrary dimensions. 
Specifically, is there a "slick" way to do this, like showing a continuous mapping from another compact set? I am allowed to use the fact that a unit sphere in arbitrary dimensions is compact, but not sure how to use that here.
Thank you!
 A: If we have a sphere $x_1^2 + x_2^2+ \ldots + x_n^2 = a$ (which we can prove to be closed easily by constructing open balls around all points in the complement), then take the continuous map $f(x_1, \ldots, x_n) = (x_1^2, \ldots, x_n^2)$, which maps the entire sphere to the plane $y_1+\ldots+y_n=a$. This is because if we have a point $(x_1, \ldots, x_n)$ on the sphere, then the function takes this point to $(x_1^2, \ldots, x_n^2)$, but $x_1^2 + \ldots + x_n^2 = a$ which means that $(x_1^2, \ldots, x_n^2)$ is on the plane. 
As shown, we just took any point on the sphere and mapped it (using a continuous function) to plane in question. Furthermore, every point in the desired portion of the plane is reached because if we have $(y_1, \ldots, y_n)$ in the desired portion of the plane, then we know that all the $y_i\geq 0$, and so taking square roots makes sense, and we know the point $(\sqrt{y_1}, \ldots, \sqrt{y_n})$ is on the sphere.
Thus the map is continuous and surjective (though not injective), and because continuous functions take compact sets to compact sets, the desired portion of the plane is compact.
