Is the set made up of all points that satisfy $g_{1}(x, y, z)=x^2 + 2y^2=1 $, $\:g_{2}(x, y, z)= xy + xz=2$ compact?

Question

I want to find the least value of a function whose domain is made up of all points that satisfy the following:

$g_1(x, y, z)= x^2 + 2y^2=1$

$g_2(x, y, z)= xy + xz=2$

In my book it is written that "a minimum exists since the domain is a closed set". However, previously there is a theorem that one can only be certain that function obtains a max and min value if the domain is compact, i.e. both closed and bounded. From what I see, the domain is not bounded, how can we then be certain that "a minimum exists"?

Any help would be appreciated.

$\textbf{Edit:}$: I want to find the least distance form the origin, so my function is $f(x, y,z)=^2+^2+^2$

Answer 1

Given the additional information, set forth in your comment, that the function you're trying to minimize is $f(x, y, z)=x^2+y^2+z^2$, the solution is easy. Just note that because $(1, 0, 2)$ is in your solution set, the minimum must fall within the (compact) sphere of radius, say, $953.8$ (or pick your favorite number that's at least as large as $\sqrt 5$). Since the solution sets of $g_1$ and $g_2$ are closed, their intersection with this compact sphere must be compact and the minimum must occur somewhere in that intersection.

Answer 2

Take $g=(g_1,g_2):\mathbb{R}^3 \longrightarrow \mathbb{R}^2$. Your set is clearly closed since $g$ is continuous and $$\{x^2+2y^2=1\}\cap\{xy+xz=2\}\cap \mathbb{R}^3=g^{-1}(1,2)$$ However, $g^{-1}(1,2)$ is not bounded. To see this, define $\vec{r}:A \longrightarrow \mathbb{R}^3 $ by $$\vec{r}(t)=\bigg(\cos(t),\frac{\sin(t)}{\sqrt{2}},2\sec(t)-\frac{\sin(t)}{\sqrt{2}}\bigg)$$ where $A=\big[0,\frac{\pi}{2}\big)$. You can check that $\vec{r}(A)\subseteq g^{-1}(1,2)$. Moreover, $$||\vec{r}(t)||\geq \Bigg|2\sec(t)-\frac{\sin(t)}{\sqrt{2}}\Bigg|\rightarrow \infty$$ as $t\rightarrow \frac{\pi}{2}^{-}$ so $g^{-1}(1,2)$ isn't bounded.