Can a function with just one point in its domain be continuous? For example if my function is $f:\{1\}\longrightarrow \mathbb{R}$ such that $f(1)=1$.
I have the next context:
1) According to the definition given in Spivak's book and also in wikipedia, since $\lim_{x\to1}f$ doesn't exist because $1$ is not an accumulation point, then the function is not continuous at $1$ (Otherwise it should be $\lim_{x\to 1}f=f(1)$). 
2) According to this answer , as far as I can understand a function is continuous at an isolated point.
I don't understand. 
Edit:


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*Spivak's definition of limit: The function $f$ approaches to $l$ near $a$ means $\forall \epsilon > 0 \; \exists \delta > 0 \; \forall x \; [0<|x-a|<\delta\implies |f(x)-l|<\epsilon]$

*Spivak's definition of continuity: The function $f$ is continuous at $a$ if $\lim_{x\to a}f(x)=f(a)$

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 $$\forall\epsilon>0,\exists\delta>0:\forall x\color{red}{\in\operatorname{dom}f},\:\bigl[0<|x-a|<\delta\implies |f(x)-l|<\epsilon\bigr]$$ Note that this definition also varies subtly and critically from Spivak's.
