Can a function with just one point in its domain be continuous?

Solution 1:

Based on the definitions Spivak gave, I suspect that (as discussed in comments) his definition of continuity is based on the assumption that we're dealing with functions defined everywhere, or at very least having domains with no isolated points. His definition does indeed break down (badly) for functions such as yours.

A related (but more general) definition given for continuity at a point $a$ of the domain of a function $f$ is something like $$\forall\epsilon>0\:\exists\delta>0\:\forall x\in\operatorname{dom}f\:\bigl[|x-a|<\delta\implies |f(x)-f(a)|<\epsilon\bigr]$$ This is provably equivalent to:

(i) $x$ is isolated in $\operatorname{dom}f$, or

(ii) $x$ is a point of accumulation of $\operatorname{dom}(f)$ and $\lim_{y\to x}f(y)=f(x)$.

The key to the proof is that for a point of accumulation $a$ of $\operatorname{dom}f,$ we say $\lim_{x\to a}f(x)=l$ iff $$\forall\epsilon>0\:\exists\delta>0\:\forall x\in\operatorname{dom}f\:\bigl[\color{red}{0<}|x-a|<\delta\implies |f(x)-l|<\epsilon\bigr]$$ Note that this definition also varies subtly and critically from Spivak's.

Solution 2:

The whole mess stems from the unfortunate definition of continuity adopted by many authors. I'd say continuity is the primary concept, and $\lim$ is an exception handling measure. For the purposes of elementary analysis a function $f$ is continuous at $a$ if for any $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-f(a)|<\epsilon$ whenever $|x-a|<\delta$. This gives you the "chain rule for continuity" for free, whereas defining continuity via limits enforces a case analysis.