The definition of continuity and functions defined at a single point In this question, the answers say that $\lim_{x \to p} f(x) = f(p) \Longleftrightarrow  f \ \text{is continuous at} \ p$ fails if $f$ is only defined at a single point. Let us consider $f:\{1\} \rightarrow \mathbb{R}$. This is continuous at $1$, yet supposedly the standard limit definition fails. 
I do not understand why. Write $D$ for the domain of $f$.  Isn't the implication $\forall \epsilon >0 \ \exists \delta >0$ s.t $\forall x \in D$ satisfying $0<|x-1|<\delta$ the inequality $|f(x) - f(1)|<\epsilon$ holds true vacuously, since there is no $x \in D$ satisfying $0<|x-1|<\delta$ regardless of our choice of $\delta$. Of course, if we define another function, say  $g$, which has an isolated point but is defined elsewhere, then we simply choose $\delta$ sufficiently small. 

Topological Assessment 
If $f: \mathbb{R} \to \mathbb{R}$ is defined at a single point then its image must also be a single point, so that the function is defined as $f(x_0) = y_0$ for some $x_0, y_0 \in \mathbb{R}$.
I'm not sure about the answer and post this more for feedback. It seems that as a function $f: \mathbb{R} \to \mathbb{R}$ it would not be continuous, but as a function $f:[x_0] \to \mathbb{R}$ then it would be.
A single point in $\mathbb{R}$ is a closed set. If $O \subset \mathbb{R}$ is any open set that contains $y_0$ then its pre-image is the closed set $[x_0]$ which contardicts $f$ being continuous on $\mathbb{R}$. 
On the other hand, considered as a function from the domain $[x_0] \subset \mathbb{R}$ then $f:[x_0] \to \mathbb{R}$ would be continuous as $[x_0] $ is both closed and open in the subspace topology of $[x_0]$
