# Property of Compactness

1. Jun 9, 2012

### trap101

Suppose S$\subset$ℝn is compact, f: S-->R is continous, and f(x)>0 for every x $\in$S. Show that there is a number c>0 such that f(x) ≥ c for every x$\in$S.

Attempt:
Since S is contained in Rn is compact, then S is closed and bounded.
By the extreme value thm there exists values a,b that are an absolute minimum and absolute maximum respectively. Let c = f(a). Therefore by EVT f(c) ≤ f(x) in R.

Well I have the solution manual and their solution is way different to what I attempted. Is there anything right about this?

2. Jun 9, 2012

### micromass

Staff Emeritus
Your solution is correct. You might want to explain why c>0 however.

3. Jun 9, 2012

### micromass

Staff Emeritus
Also, this should be c≤f(x).

4. Jun 9, 2012

### trap101

But isn't c $\in$ S. SO wouldn't the function map c to f(c)?

Is the fact c having to be positive because if c < 0 ==> f(a) < f(c) ?

5. Jun 9, 2012

### micromass

Staff Emeritus
You defined c=f(a). So c is an element of $f(S)\subseteq \mathbb{R}$. Writing f(c) makes no sense.

Again, f(c) makes no sense.

6. Jun 9, 2012

### trap101

got it now, thanks foe the help.