Proof that a continuous function on a compact set is uniformly continuous, compact set not containing infimum I'm reviewing the proof of the following statement (Baby Rudin 4.19): Let $f$ be a continuous mapping of a compact metric space $X$ into a metric space $Y$. Then $f$ is uniformly continuous on $X$.
The general idea of the proof makes sense. For a fixed $\varepsilon&gt;0$, we look at all the $ \phi(p)$ (which would usually be denoted as $\delta$), and pick one that is small enough to work for all $p\in X$. We do this by finding a finite subcover of the open cover formed by the collection of $J(p)=N_{1/2\phi(p}(p)$, call it [the finite subcover] $\{J(p_1),\ldots,J(p_n)\} $. We then let $$ \delta=\frac{1}{2}\min\{\phi(p_1),\ldots,\phi(p_2)\}.$$ Cool solution, because we cannot take $$\delta=\frac{1}{2}\inf\limits_{p\in X}\{\phi(p)\}. $$ My question is, why can't we take the infimum over all of $X$ even when $X$ is compact? Rudin writes &quot;the inf of an infinite set of positive numbers may very well be $0$&quot;, but I cannot think of any example where this is the case for a compact $X$. Surely if $X\subset\mathbb{R}$, then $\inf_{p\in X}\{\phi(p)\}&gt;0$, as $X$ is closed and bounded, and a closed and bounded set in $\mathbb{R}$ contains its infimum.
This makes me think that any compact $X$ which is an infinite collection of positive numbers such that $\inf X=0$ must somehow lack the least-upper-bound/greatest-lower-bound property, as this is what ensures a closed and bounded set in $\mathbb{R}$ contains it infimum (Baby Rudin 2.28), and any compact $X$ is closed and bounded regardless of the metric space. So perhaps we cannot take $\inf_{p\in X}\{\phi(p)\}$ as it is not defined?
I would love anyone's advice, insight, or examples!
The relevant set here is not $X$, but the range of $\phi(p)$. Is this compact? In fact it is, because $\phi$ is a continuous function of $p$, and the continuous image of a compact set is compact. So you could prove it that way. It requires two facts: that the continuous image of a compact set is compact; and that the infimum of a compact set of strictly positive numbers is strictly positive.
But Rudin's proof just uses the definition of compactness, in a natural way. So it is simpler and more straightforward.
The relevant set here is not $X$, but the range of $\phi(p)$. Is this compact? In fact it is, because $\phi$ is a continuous function of $p$, and the continuous image of a compact set is compact. So you could prove it that way. It requires two facts: that the continuous image of a compact set is compact; and that the infimum of a compact set of strictly positive numbers is strictly positive.
But Rudin's proof just uses the definition of compactness, in a natural way. So it is simpler and more straightforward.