1. ## Finding the Inf

Suppose S ⊂ R^n is compact, f:S-->R is continuous and f(x)>0 for every x ∈ S. Show that there is a number c>0 such that f(x)≥c for every x ∈ S.
Im assuming that c is the inf in this case, since it is bounded i know it exists (this comes from the compactness) but how can I show it is there?

2. Your statement is a little bit confusing, but I think that you are asking two questions. First, you say that the infimum exists because the range is bounded (due to compactness) - that is okay. Now you want to find $x_0\in S$ so that $f(x_0)=c$. I suggest constructing a sequence $\{x_n\}$ so that $f(x_n)\to c$. Use the definition of $c$ to help you.

3. Originally Posted by calculuskid1
Suppose S ⊂ R^n is compact, f:S-->R is continuous and f(x)>0 for every x ∈ S. Show that there is a number c>0 such that f(x)≥c for every x ∈ S.
Im assuming that c is the inf in this case, since it is bounded i know it exists (this comes from the compactness) but how can I show it is there?
Alternatively, since $S$ is compact one has that $f\left(S\right)\subseteq(0,\infty)$ is compact. Consequently it's closed and thus $\inf f\left(S\right)\in f\left(S\right)\subseteq (0,\infty)$. Can you conclude from there?

4. Originally Posted by calculuskid1
Suppose S ⊂ R^n is compact, f:S-->R is continuous and f(x)>0 for every x ∈ S. Show that there is a number c>0 such that f(x)≥c for every x ∈ S.
Im assuming that c is the inf in this case, since it is bounded i know it exists (this comes from the compactness) but how can I show it is there?
"Show it is there"? Do you mean show it is positive? You titled this "Find the infimum" but you don't really need to find it, just show that it is positive.

You don't really need "compact" for the lower bound- you are already told that f(S) has 0 as a lower bound. But because f(S) is compact, it is also closed- it contains all of its limit points and so contains its infimum- which must be postive because f(S) only contains positive numbers.