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. 
 A: This depends on precisely how you phrase the definition of a limit.  If you define the entire expression $$\lim_{x\to p}f(x)=L$$ as a single unit, then you are correct that if $p$ is an isolated point of the domain of $f$, this is true vacuously for every value of $L$.  On the other hand, if you define $$\lim_{x\to p}f(x)$$ to refer to the (unique) number which satisfies the $\epsilon$-$\delta$ condition, then the limit does not exist if $p$ is isolated, since the definition is not satisfied by a unique number.
I don't know which definition is more commonly used in calculus books (or whether they even pick one version unambiguously), but I would consider the single unit definition to be the "correct" definition, since as you point out it eliminates the issue with continuity at isolated points.
A: The answer is that when limits and continuity are correctly defined in a way that is applicable to this case, a function defined at a single point is always continuous. The typical calculus-book definition of limits needs to be revised, or else you can't generalize it. The difficulty is that the condition $0 < |x-1| < \delta$ should really be replaced with $|x - 1| < \delta$ for everything to work smoothly. The correct theoretical approach to relating limits and continuity is as follows. 
Let $E$ and $F$ be sets. We will consider a function $f \colon E_1 \to F$, where $E_1$ is some subset of $E$. At the calculus level, $E$ and $F$ will be subsets of $\mathbf{R}$, but in general they could be arbitrary metric spaces (or even topological spaces, although I won't discuss that case). For simplicity, I will write $|x - y|$ for the distance between $x$ and $y$, instead of $d(x,y)$ as I would in a metric space.
Let $A$ be a subset of $E_1$ and let $p \in E$ be a point in the closure of $A$ (i.e., an "adherent point" of $A$). This means that for every $\epsilon > 0$, there is some $a \in A$ with $|a - p| < \epsilon$. (If $p \in A$, then this is always the case since we can take $a = p$.)
We define the notation $\lim_{x \to p,  x \in A} f(x) = L$ to mean that for every $\epsilon > 0$, there exists $\delta > 0$ such that whenever $x \in A$ and $|x - p| < \delta$, we have $|f(x) - L| < \epsilon$. The assumption on $p$ is enough to imply the uniqueness of $L$.
For example, if $f \colon \mathbf{R} \to \mathbf{R}$, then the notation $\lim_{x \to 0, x \in (0,+\infty)} f(x)$ is simplified to $\lim_{x \to 0, x>0}f(x)$ or even, frequently, $\lim_{x \to 0^{+}} f(x)$. We can also have notations like $\lim_{x \to 0, \text{$x$ rational}} f(x)$.
The notation $\lim_{x \to p} f(x)$ on its own should, in principle, mean $\lim_{x \to p, x\in E_1} f(x)$, where $E_1$ is the entire domain of definition of $f$. In particular, if $p \in E_1$, the existence of $\lim_{x \to p} f(x)$ automatically implies that this limit is equal to $f(p)$. We can then simply define $f$ to be continuous at a point $p \in E_1$ whenever $\lim_{x \to p} f(x)$ exists.
But most often, in calculus books (or in more advanced books when the distinction is unimportant), $\lim_{x \to p} f(x)$ is taken to mean what we would technically write as $\lim_{x \to p, x \ne p} f(x)$. The shorter notation is advantageous in some ways because limits of the form $\lim_{x \to p, x \ne p} f(x)$ are considered so frequently. 
However, from a theoretical perspective, considering the condition $x \ne p$ to be implied is quite inconvenient. For example, it is this convention that makes "composition of limits" fail. That is, we can have $\lim_{x \to p, x \ne p} f(x) = q$ and $\lim_{y \to q, y \ne q} g(y) = r$ without necessarily having $\lim_{x \to p, x \ne p} g(f(x)) = r$. (Take $f(x) = x \sin 1/x$, $g(y) = 0$ for $y \ne 0$, $g(0) = 1$, $p = q = 0$.) On the other hand, it is entirely true that if $f(A) \subseteq B$, $\lim_{x \to p, x \in A} f(x) = q$ and $\lim_{y \to q, y \in B} g(y) = r$, then we must have $\lim_{x \to p, x \in A} g(f(x)) = r$.
To come back to your original question, what happens if $E_1 = \{p\}$? In this case it is easy to see that the limit $\lim_{x \to p} f(x)$, which means (or ought to mean) $\lim_{x \to p, x \in \{p\}} f(x)$, always exists and has the value $f(p)$. If you apply the calculus-book definition of a limit, it will denote $\lim_{x \to p, x \in \varnothing} f(x)$, which is not meaningful because $p$ is not an adherent point of $\varnothing$.
Let me note in closing that the systematic use of notations like $\lim_{x \to p, x \ne p} f(x)$ is not unheard of even at the calculus level. This approach was standard in high schools in France for many years, where $\lim_{x \to p} f(x)$ was defined the way I've defined it here. I haven't ever seen it done this way in English-language calculus textbooks, however.
Edit I've had a look at Apostol's analysis book, and the definition of a limit given there also assumes $x \ne p$ implicitly. In that case $p$ is required to be an "accumulation point" of $A$, that is, an adherent point of $A - \{p\}$. One drawback to this approach (in addition to the composition issue I've already mentioned) is that he is required to give separate definitions of limits and continuity, rather than using limits to define continuity.
A: 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 $[x_0]$
