Extension of a compact set via continuous functions

Question

Define a set $\mathcal{D}\subset \mathbb{R}^n$ and a continuously differentiable function $h:\mathcal{D} \to \mathbb{R}$. Consider the set $\mathcal{C} = \{ x \in \mathcal{D} : h(x) \geq 0\}$. Let $h$ be defined on $\mathcal{D}$ such that $\mathcal{C} \subset \mathcal{D}$, and is compact.

Now consider the following set: $$ \mathcal{C}_e = \{ x \in \mathcal{D} : h(x) \geq - \epsilon \} \subset \mathcal{D}, $$ for some $\epsilon >0$.

What can we say about $\mathcal{C}_e$. Is it compact as well?

score 0 · Answer 1 · answered Feb 20 '18 at 15:25

0

$C_e$ is closed since it is the primage of $h^{-1}[\epsilon,\infty)$ and a closed subset of a compact set is again compact (in this case, since it is bounded.)

answered Feb 20 '18 at 15:25

Andres Mejia

20,977

I see. The set is $h^{-1}([-\epsilon,\infty))$, which is closed. $$ h^{-1}([-\epsilon,\infty)) = h^{-1}([-\epsilon,0]) \cup h^{-1}([0,\infty)) = h^{-1}([-\epsilon,0]) \cup \mathcal{C} $$ – Shishir Feb 21 '18 at 02:02

score 0 · Accepted Answer · answered Mar 09 '18 at 04:40

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It is not true. Here is a counter example.

Counter example for the compactness arguement

answered Mar 09 '18 at 04:40

Shishir

25
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You get obvious counter-examples with $\mathcal C$ is empty. – Kavi Rama Murthy Mar 09 '18 at 06:03
Excellent point, empty set is compact as well. – Shishir Mar 16 '18 at 05:58

Extension of a compact set via continuous functions

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