Suppose we have a monotone sequence of sets: $$A_1,\ldots,A_n$$ $$A_i \subseteq A_{i+1}$$

I think this is a function from $\mathbb N$ to a space of sets. Can we define a function from $\mathbb R$ to a space of sets? Could we then define a derivative of this function?

  • $\begingroup$ Could we define the derivative of a set-valued function $A:\mathbb R\to 2^X$? Which kind of arithmetic in $2^X$ is on your mind? $\endgroup$
    – user251257
    Commented Jul 23, 2015 at 17:41
  • $\begingroup$ What exactly do you mean by a derivative? Something analogous to the derivative from differential calculus? $\endgroup$ Commented Jul 23, 2015 at 17:41
  • $\begingroup$ There exist a derived set (in Polish "pochodna zbioru" i.e. derivative of the set): en.wikipedia.org/wiki/Derived_set_(mathematics) $\endgroup$ Commented Jul 23, 2015 at 17:41
  • $\begingroup$ Yes, I am thinking of something analogous to differential calculus, or, more precisely, to the first difference of a sequence (the difference between sucessive members). I was just wondering if there are already readily defined concepts for this. $\endgroup$
    – blue-dino
    Commented Jul 27, 2015 at 21:47

3 Answers 3


Great question!

You're absolutely right that we can think of a sequence of (not necessarily nested) sets as a function from $\mathbb{N}$ to $\textbf{Sets}$, and that - to the extent that we can think about $\mathbb{N}$ and $\textbf{Sets}$ as spaces - we might have a notion of derivative. However, it turns out that $\mathbb{N}$ is too bad for this to work nicely.

In order to make sense of the "derivative" of a function from $X$ to $Y$, we need $X$ and $Y$ to both carry a notion of distance - that is, we need $X$ and $Y$ to be metric spaces. On the face of it, this is then enough to define a difference quotient (EDIT: as lisyarus pointed out, this does not generalize the derivative, but rather the absolute value of the derivative: without a notion of direction, it is impossible to tell whether a function is "increasing" or "decreasing," etc.) as follows: $f'(a)$ is the limit, as $b$ approaches $a$, of the ratio $${d_Y(f(a), f(b))\over d_X(a, b)}$$ (where $d_X$ and $d_Y$ are the notions of distance on $X$ and $Y$, respectively). However, there is a huge problem with this: we're assuming that this limit, if it exists, is unique. In general, this won't be the case. For example, consider the natural metric on $\mathbb{N}$ ($d(m, n)=\vert m-n\vert$). This is full of "gaps," and these gaps prevent the derivative from making sense: given any function $f$ from $\mathbb{N}$ to $Y$, any $n\in\mathbb{N}$, and any real number $r$, we have $$\forall \epsilon>0\exists \delta>0\forall k[0<\vert k-n\vert<\delta\implies \vert{d_Y(f(a), f(b))\over \vert k-n\vert}- r\vert<\epsilon]$$ for stupid reasons: take $\delta<1$. So the statement "the difference quotient tends to $r$," as naively written, will be true for all $r$.

Basically, in order to have a useful notion of derivative of a function from metric space $X$ to metric space $Y$, we need $X$ to have no isolated points. If there are no isolated points in $X$, then the limit of the difference quotient (if it exists) is unique, so the theory of differentiation isn't totally broken. Of course, without further niceness assumptions on both $X$ and $Y$ it probably won't be good . . .

In case you're interested in crazy examples of this sort of thing, which are actually mathematically useful:

Most of the time you'll hear about calculus being generalized to manifolds; these are spaces that "look like" $\mathbb{R}$ in some sense, so the existence of such a generalization isn't really very surprising. For an example of calculus done on a wildly different space than $\mathbb{R}$, check out p-adic calculus: http://www2.math.ethz.ch/education/bachelor/seminars/hs2011/p-adic/report8.pdf


To begin with, there is no "space of sets". Even a "set of all sets" does not exist (in ZFC and its extensions, for example). Your functions should have values in some set, whose elements are sets themselves.

$\mathbb{N}$ and $\mathbb{R}$ are sets also, and nothing prevents you from speaking of a function from them to somewhere with values being sets. For example $f: \mathbb{R} \to 2^\mathbb{R}$, $f(x) = \{y : y < x\}$.

To talk about derivatives, you have to define a structure of a Banach space on the function's codomain (that is, roughly, you have to define what do addition, multiplications by scalar, size and limits mean for your sets). If you do this, then you can define a derivative.

  • $\begingroup$ You don't really need the whole Banach space structure (although of course it makes things nice), it's enough to have a metric structure with no isolated points; see my answer. $\endgroup$ Commented Jul 23, 2015 at 17:55
  • 3
    $\begingroup$ @NoahSchweber I thought about your answer, and two things made me doubt. First, usually, the derivative has the same "type" as the function itself: the derivative of $f:\mathbb{R}\to\mathbb{R}^n$ is again a function $f':\mathbb{R}\to\mathbb{R}^n$, and in your example the derivative is always $f':\mathbb{R}\to\mathbb{R}$. Also, your one is always positive, which shows that it is highly different from normal derivative. Secondly, for a suitable bad-behaved function your derivative will depend on actual path of $b$ approaching $a$, and the limit may not exist at all. $\endgroup$
    – lisyarus
    Commented Jul 23, 2015 at 18:01

Since you have a finite sequence of sets, a discrete derivative would be appropriate. For a function $f:\mathbb Z\to\mathbb R$ the discrete derivative is $\Delta f(x)=f(x+1)-f(x)$.

For a function $f:\mathbb Z\to\cal{P}(X)$ (the power set of any set $X$) you could define $\Delta f(n)=f(n+1)\setminus f(n)$. If $f(n)\subset f(n+1)$ for all $n$, this would look like a natural definition of a discrete derivative.

You could also try to define a non-discrete derivative, but it is going to be a different kind of object. For a set $A\subset\mathbb R^n$, let $\chi_A$ be it's characteristic function. This characteristic function can be regarded as a measure (or a distribution). Now for a function $f:\mathbb R\to\cal{P}(\mathbb R^n)$ you could define the derivative as $$ f'(x)=\lim_{h\to0+}\frac1h(\chi_{f(x+h)}-\chi_{f(x)}) $$ if the limit exists. The limit is taken as a measure or a distribution, not as a function or a set.

For example, if $\eta:\mathbb R\to\mathbb R$ is smooth and $n=1$, we can look at the set-valued function $f(x)=(-\infty,\eta(x))$. Then the derivative is the distribution (or signed measure) $f'(x)=\eta'(x)\delta_{\eta(x)}$.

  • $\begingroup$ Ifthat limit exists it is zero. $\endgroup$ Commented Jul 23, 2015 at 17:51
  • $\begingroup$ @MarianoSuárez-Alvarez, how so? If $f(x)=[0,x]\subset\mathbb R^1$, then $f'(x)=\delta_x$ (for $x>0$) and similar things can happen in higher dimensions. $\endgroup$ Commented Jul 23, 2015 at 17:54
  • $\begingroup$ Once you fix $x$, the limit is that of a function which takes values in $\{-1,0,1\}$. $\endgroup$ Commented Jul 23, 2015 at 18:00
  • $\begingroup$ @MarianoSuárez-Alvarez, the limit is taken as a (signed) measure or as a distribution, not as a function. I edited to make this more explicit. $\endgroup$ Commented Jul 23, 2015 at 18:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .