Given two closed sets $A$ and $B$ living in topological spaces $X$ and $Y$, the boundary of $A\times B$ in the product topology, denoted (suggestively) by $\partial(A\times B)$, is given by

$$\partial(A\times B) = ((\partial A)\times B)\cup (A\times(\partial B)).$$

For three closed sets, we have

$$\partial(A\times B\times C) = ((\partial A)\times B\times C)\cup (A\times(\partial B)\times C) \cup (A\times B\times(\partial C))$$

and so on. This is extremely similar to the product rule for derivatives, with $\times$ replaced by $\cdot$, $\partial$ replaced with the derivative and $\cup$ replaced with $+$. This seems way too similar to be a happy coincidence but my imagination can't seem to easily make a connection between derivatives and boundaries at an fundamental level. There are quite obvious connections between the two at the level of smooth manifolds but I'm not sure if that is a very satisfying connection since it seems to put the cart before the horse. I'm aware of such fields as differential algebra which study algebraic structures endowed with objects which obey something resembling the Leibniz law and this seems to be along the lines of this phenomenon. Is this truly a coincidence or should there be deep reasons for why the product rule underlies both derivatives and boundaries of (closed) sets?

  • $\begingroup$ What is the obvious connection at the level of smooth manifolds? $\endgroup$
    – JonHerman
    Sep 21, 2014 at 0:47
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    $\begingroup$ What I had in mind was Stokes' theorem since it naturally connects boundaries with derivatives (more specifically differential forms). $\endgroup$ Sep 21, 2014 at 0:50
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    $\begingroup$ Does this answer your question? mathoverflow.net/questions/46252/… $\endgroup$
    – user157227
    Oct 19, 2014 at 3:09
  • $\begingroup$ That seems like a nice use of the bounty points I awarded you! ;-) $\endgroup$
    – Asaf Karagila
    Oct 19, 2014 at 3:34
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    $\begingroup$ More generally, $\partial(A\times B)=(\partial A\times\overline B)\cup (\overline A\times \partial B)$ $\endgroup$
    – Pedro
    Oct 19, 2014 at 3:42

2 Answers 2


I think what would be fruitful would be to consider this in terms of measure theory; then, we can actually formalize Robin's answer. We want to create a structure in which we can embody both sets and scalars as the same sort of object. A convenient way to do this is by using measures. Later in the post, I'll get to how this connects back, but first let's set up some of the machinery we need to create a "boundary" on measures.

In particular, consider signed measures $\mu_1$ and $\mu_2$ defined on sets $S_1$ and $S_2$. Essentially, you can think of these as just a generalized notion of a function or a distribution. We can define a product measure $\mu=\mu_1\times \mu_2$ on $S_1\times S_2$ by defining that, for $A_1\subseteq S_1$ and $A_2\subseteq S_2$, the product measure satisfies $\mu(A_1\times A_2)=\mu_1(A_1)\mu_2(A_2)$. A convenient notion about measures is that we can multiply them by scalars - i.e. $(c\cdot \mu)(A)=c\cdot\mu(A)$, and we can take differences of them*. Moreover, we can define limits of them such that $(\lim_{t\rightarrow\infty}\mu_t)(A)=\lim_{t\rightarrow\infty}(\mu_t(A))$ where $\mu_t$ is a family of measures parameterized in $t$.

In particular, this means that if we have some family of measures $\mu_t$, we can define it's derivative with respect to $t$ as: $$\mu'_t=\lim_{h\rightarrow 0}\frac{1}h(\mu_{t+h}-\mu_t).$$ We actually have every tool already that we need to show that this derivative satisfies the product rule - you can prove it just the same way you'd prove it for real valued functions, since the measures with $\times$ and $+$ act basically the same as real numbers. Further, since if we had two measures $\mu_t$ and $\nu_t$ and wanted to know about the derivative of $\mu_t\times \nu_t$ we would only have to consider its value over some basis, we could see that, since from definition $(\mu_t\times\nu_t)(A_1\times A_2)=\mu_t(A_1)\nu_t(A_2)$, our limit is going to have to obey the product rule, since we basically have a product of two real valued function. Moreover, if we think of the derivative as how fast the measure is "expanding", the only terms that are expanding linearly are the terms where $\mu_t$ is expanding, but $\nu_t$ is constant and vice versa, since the ones where both are expanding would be quadratic, and eliminated in the the limit.

Now, we get to the bit where some of this abstract nonsense begins to work its magic - don't fret too much if the second paragraph didn't make much sense. It's not as important as what's next. First, we will relate this new measure derivative to the standard one. In particular, let $f(t)$ be some differentiable function $\mathbb{R\rightarrow R}$. We are going to relate it to a family of measures $\mu_t$ on the set $\{s\}$ by setting $\mu_t(\{s\})=f(t)$. It should be quite obvious that $\mu_t'(\{s\})=f'(t)$ from our definition. Moreover, if we had two such functions, $f$ and $g$ and corresponding families of measures $\mu_t$ and $\nu_t$ on singleton sets $S_1$ and $S_2$, we would have $$(\mu_t\times \nu_t)(S_1\times S_2)=f(t)g(t)$$ and hence $$(\mu_t\times \nu_t)'(S_1\times S_2)=(fg)'(t)$$ and if we expand the right side by the product law for measures, we get $$\mu_t'(S_1)\nu_t(S_2)+\mu_t(S_1)\nu_t'(S_2)=(fg)'(t)$$ where the left hand terms can be read as $f'(t)g(t)+f(t)g'(t)$ - meaning that measures satisfying the product law implies that so do functions $\mathbb{R}\rightarrow\mathbb{R}$.

Now, the real trick is to relate this to sets. To do this, we are going to choose some canonical measure (e.g. the one that gives the volume of a set) $m$ on a set $S$ and we will define, for any set $R\subseteq S$ the measure $\hat{R}$ as $$\hat{R}(A)=m(A\cap R).$$ Intuitively, this measure says, "measure the volume of $A$ within the region $R$". So, for instance, if $R$ were the interval $[0,1]$, then the measure would be the length of $A$ within $[0,1]$, ignoring any parts outside. As an example, suppose we let $R_t=[0,2t]$. Then, we wish to calculate the derivative of the corresponding measure $\hat{R}'_t$. We can see that, since, for a closed set $A$, the value $\hat{R}_t(A)$ and that of $\hat{R}_{t+h}(A)$ agree for small enough $h$ if $A$ does not contain $2t$. If $A$ contains $2t$ on its interior, then the difference $\hat{R}_{t+h}(A)-\hat{R}_{t}(A)=m([2t,2t+2h]\cap A)$ which equals, for small enough $h$, exactly $2h$. Hence, $\hat{R}_{t+h}(A)$ is $2$ if $A$ contains $2t$ and $0$ otherwise. We can interpret this to mean that the family of sets $R_t$ does not vary with $t$ at most points, except around the point $2t$, where, as $t$ increases, $R_t$ grows from that point at a rate of $2$. You can similarly visualize this as a rectangular pulse of width $2t$ which is growing at a rate of 2. The derivative of the measure represents how quickly it changes at a given time, and hence will only have non-zero values for sets intersecting the boundary, as others do not change.

However, even this doesn't quite get us to where we want; we need to, given a set $R$, define a family $R_t$ and take its derivative. Borrowing Robin's equation, we let $R_t$ be the $t$-neighborhood around $R$. So, we expect to find that the derivative $\hat{R}_0'$ will be a measure telling us the boundary of $R$, because we are making it "expand" at a rate of $1$. In particular, if $m$ measured volume, we can show that $\hat{R}_0'(A)$ is equal to the surface area of the boundary of $R$ within the interior of $A$. To see this, just notice that, as $R_h$ goes to zero, the difference $R_h-R_0$ is just a thin "film" running around the boundary with thickness of about $h$ at every point. Thus, its volume is roughly the surface area of $R$ times $h$ - and the derivative cancels out the "times $h$" term. Note that, from the derivative measure, we can extract the boundary just as the set of points $p$ such that for any neighborhood $A$ of $p$ the value $\mu(A)$ is not zero.

The next insight to make is that, if we take two sets $R$ and $T$, then the measure corresponding to $R\times T$ is the same as the measure as $\hat{R}\times \hat{T}$. Therefore, if we try to find the derivative of the family of measures associated $R\times T$, we are actually finding the derivative of $\hat{R}_t\times \hat{T}_t$, which, since measures will satisfy the product rule, is $\hat{R}_0'\times \hat{T} \cup \hat{R}\times \hat{T}_0'$ - and sense, when we extract the boundary from this measure, we are going to end up with $\partial R \times T \cup R\times \partial T$, sense we previously showed that $\hat{R}_0'$ represents $\partial R$ and taking a product of measures like this will take a product of the corresponding "non-zero" elements of the measure, this establishes that the product rule for measures also implies it for the boundary of sets - meaning that we can interpret the fact that this equality holds for boundaries and derivatives as coming from a common fact that either can be expressed in terms of a derivative of a measure.

I hope this actually helps link these concepts in your mind - this explanation helps me see the link, but it feels a bit too long to truly be intuitive (especially if you haven't internalized the notion of a measure), and is rather abstract in any case. I'd be happy to try to clarify anything.

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    $\begingroup$ This is one of the coolest answers I've seen on this site. I never would have made the connection with measures but with the way you described it, it makes perfect sense. This is exactly the kind of answer I was looking for. $\endgroup$ Oct 19, 2014 at 14:08
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    $\begingroup$ Seeing this for the first time 4 years after its posting, this is phenomenal answer. $\endgroup$ Jul 27, 2018 at 22:48

Here's an informal definition:

Suppose $X$ is a metric space and $A$ is a subspace of $X$. Let $A_t$ be the $t$-neighborhood of $A$. Then, $$\partial A = \lim_{\hspace{.6em}t\rightarrow 0^+}\frac{1}{t}(A_t\setminus A).$$

For full disclosure, I thought I had first seen this on Stack Exchange, but a quick search didn't turn up anything. Regardless, this is how I first understood precisely why they occasionally act similarly. Of course, the above definition is completely informal, and I have yet to find a pain-free way of making it rigorous, but it gets the point across: the $t$-neighborhoods, when we remove $A$, approach the boundary.

There's more too, in the area of differential topology, with de Rham's theorem being of particular interest. The infamous de Rham's theorem connects regular cohomology, using boundaries of cochains, to de Rham cohomology, using the exterior derivatives of differential forms, by noting that they give essentially the same information.

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    $\begingroup$ How can you divide by a number in a general metric space? This is pretty neat though. $\endgroup$ Sep 21, 2014 at 14:44
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    $\begingroup$ You're not really doing that. The formula just gives a mental picture of how the boundary operator can act like a derivative. It's only intuition. $\endgroup$ Sep 21, 2014 at 15:09
  • $\begingroup$ Infamous de Rham's theorem. How is that? $\endgroup$
    – Thomas
    Nov 29, 2022 at 19:48

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