formal differences in definition between calculus and analysis?

Question

I've noticed some differences in definitions between lower level math books and analysis books:

1. In calculus style limits, the function has to be defined in a deleted open interval of the point. But this is not a strict requirement in analysis. For example, it makes sense to talk about $\lim_{x\to 0}\sqrt{x}$ (and this is obviously a one-sided limit, but you don't necessarily have to call it that).

2. A lot of calculus books also require functions be defined in a deleted open interval of a point of continuity so that "continuity at a point" is equivalent to the usual limit statement $\lim_{x\to p}f(x)=f(p)$. In analysis, continuity is defined more generally.

3. The phrase "continuous on an interval $[a,b]$" actually means something different in calculus than it does in analysis. The function $f(x)=1$ when $x\geq 0$ and $f(x)=0$ when $x<0$ is continuous on $[0,1]$ in the calculus sense since we only require one sided continuity at the endpoint $0$. In analysis, "$f$ is continuous on $[0,1]$" would not be true. It seems calculus textbooks are talking about continuity with respect to the function $f$ restricted to the interval $[a,b]$.

4. The function $1/x$ is sometimes considered to have a discontinuity at $0$. But in analysis, points of discontinuity need to be part of the domain. So actually $1/x$ is continuous everywhere (in its domain).

What are some other differences? Is there a list somewhere?

Answer 1

In reality there are no differences. Fact is that in the "calculus environment" one does not elaborate on all subtleties of the fine print. Now you are looking at the fine print.

1. If "the point" is $\xi\in{\mathbb R}$, and we are looking at limits $x\to\xi$ the function $f$ has to be defined in points arbitrarily near $\xi$. If, e.g., $f$ is defined for $\xi<x<b$ for some $b>\xi$ then it makes sense to look at $\lim_{x\to\xi} f(x)$ without talking about "relative topology".

2. The function $g:\>x\to{1\over x}$ is undefined at $x=0$, and this function is continuous (even $C^\infty$) wherever it is defined. Even in a "calculus environment" one should not say that $g$ has a discontinuity at $x=0$. What we see there is a disconnectivity of the domain. Note that we cannot create a continuous function $\tilde g$ on ${\mathbb R}$ by defining $\tilde g(0)$ suitably, as in the case $x\mapsto{\sin x\over x}$.