# How to prove the sublevel set is compact?

Let $x_1$ and $x_2$ be two vectors in $\mathbb{R}^2$. Consider an objective function $$V(x)=(\|x_1\|^2-d_1^2)^2 + (\|x_2\|^2-d_2^2)^2$$ where $d_1$ and $d_2$ are two positive constants, and the vector $x=[x_1^T,x_2^T]^T\in\mathbb{R}^4$. Define the sublevel set $$\Omega_c=\{x\in\mathbb{R}^4 \ |\ V(x)\le c\}$$ Can anyone show how to prove this set $\Omega_c$ is compact (closed and bounded)? Thanks.

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Hint: For closedness check continuity of $V$, for boundedness, check what happens to $V(x)$ for $\|x\|\to\infty$.
For boundedness, if $\|x\|\rightarrow \infty$, $V(x)\rightarrow \infty$. So if $V(x)\le c$, then $\|x\|$ must be bounded. But for closedness, can the continuity of $V$ imply the closedness? I'm not familar with topology or related math. Can you explain it in more detail? – Shiyu Mar 26 '12 at 6:24
Basically, if a function is continuous, then inverse images of closed sets are closed. Since $]-\infty,c]$ is a closed set, this works here. – Dirk Mar 26 '12 at 10:11
Thanks. But the function $V(x):\mathbb{R}^4\rightarrow \mathbb{R}$ is not a bijection, right? I mean the inverse of $V(x)$ does not exist. Then can we still claim that using the property you mentioned? – Shiyu Mar 26 '12 at 13:58
Inverse images of sets always exist: The inverse image of $A$ under $V$ is the set $\{x\ :\ V(x) \in A\}$. – Dirk Mar 26 '12 at 18:43