# Differentiability of a singleton set

Let $f$ be a function $f:\{2\} \rightarrow\mathbb R$. Now absolutely $f$ is continuous. But what we can say about the differentiability of the function $f$ at $x=2$ ?

The function $f$ is not differentiable at $2$ since $2$ is not an accumulation point of the domain of $f$.

• Sir, in the book of Analysis On Manifolds by Munkres, the author defines differentiability for a point $x_0$ in an arbitrary subset $S$ of $\mathbb{R}^n$ as: if we have a differentiable function $g$ defined on a neighbourhood $U$ of $x_0$ in $\mathbb{R}^n$ s.t $g$ agrees with $f$ on $U \cap S$; Can't we use this definition for the above question ?
– Our
Jul 13, 2018 at 10:49
• Note: the definition above given in the page 144, question 3.a
– Our
Jul 13, 2018 at 10:50
• @onurcanbektas Munkres defines the concept of differentiability of a function $f$ at a point $a$ at the start of §5. And he defines it only for functions whose domain contains a neighborhood of $a$. Jul 13, 2018 at 10:52
• Yes, I'm aware of that definition, but the above definition is, I guess, and extension of the that definition for sets that do not contain a neighbourhood of a point in the set. I mean, I'm directly quoting the author, and asking can't we use the above definition that the author provided later in the examples of the chapter 16.
– Our
Jul 13, 2018 at 10:56
• @onurcanbektas Yes, using that definition of differentiable function taken from that exercise, your function is differentiable. Were we supposed to have guessed that that's what you had in mind one you posted your original question? Jul 13, 2018 at 11:02

By definition, a function is differentiable at $x=2$ if the limit $\lim_{x\to\ 2}\frac{f(x)-f(2)}{x-2}$ exists. But in this case, since $2$ is the only value of $x$ for which $f(x)$ is defined, and $x-2=0$ for that value of $x$, the expression $\frac{f(x)-f(2)}{x-2}$ is not defined for any $x$. Therefore, the limit cannot exist.

From the viewpoint of smooth manifold, you treat $$\{2\}$$ as a $$0$$-dimensional manifold. Any function on a $$0$$-dimensional manifold is automatically differentiable. But the derivative would not be a slope of some tangent line, but rather the uninteresting zero linear map $$T:\{\vec0\}\to\Bbb R$$, $$T(\vec0)=0$$.