Def: A function $f$ from a metric space $X$ to $Y$ is uniformly continuous if given $\epsilon >0$ there exists a $\delta >0$ s.t for $x,y\in X, D(x,y)<\delta \to D(f(x),f(y))< \epsilon$.
Here is the quote from Sets and Metric spaces by Kaplansky, Pg 102
The idea behind this def. is that the epsilon-delta def. of continuity is applied globally, rather than one point at a time...Let us give an example to illustrate the point. As we shall see in Section 5.3, no such example can be given with $X$ a closed interval - or any bounded closed set.
My understanding of this quote Kaplansky seems to be saying if we are dealing in $\mathbb R$ and some function $f$ on $X=[a,b]$ then $f$ is not uniformly continuous?
Isn’t this def pretty much the same intro analysis courses. Then the basic assumptions should hold for metric spaces as continuity does?
As far the example went to section 5.3 it is on compactness. The only Theorems it could apply is 73 or 67? But these are theorems and not examples per say..
I don't have the book, nor a PDF of it. But probably he gives an example like $f(x)=\frac{1}{x}$ on $(0,1)$ to have a function that is continuous at each point but not uniformly continuous on its domain. $f(x)=x^2$ also works as such an example on $\Bbb R$ (or any unbounded real set), say (all in the standard metrics of course).
Later he will presumably prove the theorem that any continuous (i.e. normal, pointwise continuous) $f$ on a space $X$ that is compact is automatically uniformly continuous, a very useful and classic fact in analysis. So the examples he must ahve given earlier had a non-closed or non-bounded domain (in $\Bbb R$) as in the reals compactness can be shown to be equivalent to being closed and bounded.