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Let $X$ a topological space, A ⊆ X compact and $f : A → \mathbb{R}$ a continuous function with the property that $f(a) > 0$, $∀a ∈ A$. Show that there is a $c > 0$ such that $f(a) ≥ c, ∀a ∈ A$. I thought to use the fact that if $A$ is compact then $f$ is bounded and $f(A)$ is closed, so there is $x_1\in A$ such that $f(x_1)=\inf f(X)$ and if I know that for any $a\in A$ the function is bigger than zero, then $\inf f(x) \geq 0$. It seems to simple to be true however.

More on the result I used here $X$ compact metric space, $f:X\rightarrow\mathbb{R}$ continuous attains max/min

Did I give a complete proof?

bof
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1 Answers1

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There are two problems here:

  1. Yes, $f(A)$ is closed, but you should explain why. It is closed because it is compact and it is compact because $f$ is continuous and $A$ is compact.
  2. The fact that $\inf_{x\in A}f(x)\geqslant0$ is irrelevant here. What matters is that, since $f(A)$ is closed, $\inf f(A)\in f(A)$. So, there is some $a_0\in A$ such that $\inf f(A)=f(a_0)$, and then, for each $a\in A$, $f(a)\geqslant f(a_0)>0$. So, take $c=f(a_0)$.
José Carlos Santos
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