Why does a global extremum on an open interval not exist? Say for example the function $f(x)=x^2$ on the interval $(0,\infty)$. I was taught that there is no global minimum on this interval. But I still can't quite wrap my head around why. $\lim_{x\to0^+}f(x)=0$, so there is a point infinitesimally close to $x=0$ that is smaller than every other value in the range of the function on the interval. Why then isn't that value considered the global minimum? I've always taken it to be a "by definition" thing, that the definition of absolute extremum simply won't allow it, but is there a more analytical argument for this?
 A: The issue is in the claim "there is a point infinitesimally close" - in standard analysis there is no such thing as an infinitesimal. The domain of the function is the set of real numbers strictly greater than zero. For any such $x$, $x/2$ is also in the domain and has $f(x/2) < f(x)$, which proves there is no minimum on the interval.
A: The real numbers possess a remarkable property called completeness.  It means that the limit of any sequence of real numbers is, again, a real number.
Hence, if you take $x$ values closer and closer to 0, their squares ($f(x)$) will get closer and closer to 0.  The limit of their squares must be a real number, say $L$, by the completeness property.  
However real numbers cannot be infinitesimally close to other real numbers, because the difference between two real numbers is a real number.  Hence the difference between $L$ and $0$, which must be a real number, can be either 0 or nonzero.  If it is 0, then there is a problem because $f(x)$ isn't 0 on $(0,\infty)$.  If it is nonzero, then again there is a problem because for small enough $x$, $f(x)$ will be less than $L$ so that can't be a limit.
