I need to prove that the following set: $$\mathcal L:=\{x\in \mathbb{R}^n \, : \, f(x) \leq f(x_0) \}$$ which apparently is called lower level set, is compact. I've already proved boundedness and I just need to show closeness.
The function $f$ is twice continuously differentiable on all of $\mathbb{R}^n$ and so it is continuous everywhere.
I am looking for a "simple" proof, i.e. one that uses the notion of continuity and the notion of closeness/openness, NOT the notion of semi-continuity, which is used in every single proof I've found.
My Attempt
$f$ is continuous on all of $\mathbb{R}^n$ means that $\forall x\in\mathbb{R}^n$: $$\forall \epsilon >0, \, \exists \delta > 0 \, \forall y\in \mathbb{R}^n: ||x - y|| < \delta \Longrightarrow ||f(x)-f(y)||< \epsilon$$ On the other hand, $\mathcal L$ is closed means that its complement wrt $\mathbb{R}^n$ is open, that is: $$\forall x\in\bar{\mathcal{L}}, \exists \epsilon > 0 \, \forall y\in \mathbb{R}^n \, : \, ||x - y||<\epsilon \Longrightarrow y\in \bar{\mathcal{L}}$$
Now using continuity we know: $$||x-y|| < \delta \Longrightarrow ||f(x)-f(y)||<\epsilon \Longrightarrow -\epsilon < f(x) - f(y) < \epsilon$$ And so $\Longrightarrow f(y)-\epsilon < f(x) < \epsilon+f(y)$. How do I continue? Is this correct?
If we define $$ g(x)=f(x)-f(x_0) $$ a continuous function, the set is $$ g^{-1}((-\infty,0]) $$ the preimage of a closed set under a continuous map.
If you are wedded to epsilon's and delta's, then pick $x\in \mathcal{L}^c$, then $$ g(x)>0 $$ and for $\epsilon=\frac{g(x)}{2}$, we have for some $\delta>0$ corresponding to this $\epsilon$, $$ g(y)>0 $$ for all $y\in (x-\delta,x+\delta)$. So $(x-\delta,x+\delta)\subset \mathcal{L}^c$.