Existence of minimum in bounded but open set According to the Extreme Value Theorem, a continuous function achieves at least one minimum and one maximum whenever the set is bounded and closed (i.e. compact).

In my case, I have a bounded and open set, thus, although an infimum exists, a minimum is not assured. 

But consider the extra information:


my continuous function is like $z=f(x,y)$, with $f: A \rightarrow \Bbb R$, and $A \subset \Bbb R^2$
from the plot I see that at the boundary the function is far from the minimum, and the minimum is at the interior. In other words, the infimum is not at the boundary.


The question is:

How can I formalise this graphical insight in order to apply the Extreme Value Theorem to prove that my function necessarily has a minimum? I am thinking on the following logic:


continuous function +
Bounded and open set +
Infimum not at the boundary + 
Extreme Value Theorem 


=> Minimum necessarily exist.
I would say yes.

From what you said, it seems that you can continuously extend your function $f$ on $A$ to a function $\tilde{f}$ on $\bar{A}$ (where $\bar{A}$ is the minimum closed set containing $A$) and apply the Extreme Value Theorem to $\tilde{f}$.

Similarly if you know that the infimum is not at the boundary than you know that the infimum is in $B \subset A$ with $B$ closed (and thus compact) and apply the Extreme Value Theorem to $f$ restricted to $B$.
I would say yes.

From what you said, it seems that you can continuously extend your function $f$ on $A$ to a function $\tilde{f}$ on $\bar{A}$ (where $\bar{A}$ is the minimum closed set containing $A$) and apply the Extreme Value Theorem to $\tilde{f}$.

Similarly if you know that the infimum is not at the boundary than you know that the infimum is in $B \subset A$ with $B$ closed (and thus compact) and apply the Extreme Value Theorem to $f$ restricted to $B$.