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I was reviewing the answer to this problem here and am trying to understand why condition 3 holds. That is: why does a real, continuous function in $\mathbb{R}^n $, where ()→ +∞ as ‖‖→+∞, have compact sublevel sets?

I can picture sublevel sets in a 2D plane, but I think I am getting confused when trying to interpret them in a higher dimension.

Math927
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Sep 24 '22 at 10:43
  • Do you understand what “$f(x) \to +\infty$ as $\Vert x \Vert \to +\infty$” means? – Martin R Sep 24 '22 at 11:00
  • I'm thinking about it geometrically, so as the length of my vector gets larger and larger (going to infinity), my function is also going to infinity. – Math927 Sep 24 '22 at 11:06

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A sublevel set $f^{-1}[(-\infty,c]]$ is closed since $f$ is continuous and $(-\infty,c]\subseteq\mathbb{R}$ is closed.

The condition that $\lim\limits_{\lVert x\rVert\to\infty}f(x)=\infty$ enforces that $f^{-1}[(-\infty,c]]$ is bounded for all $c\in\mathbb{R}$ (think about what happens if $f^{-1}[(-\infty,c]]$ is unbounded for some $c\in\mathbb{R}$).

Since every sublevel set is closed and bounded, then by the Heine-Borel theorem, they are compact.

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