Given the empty function $\varnothing: \emptyset \to X$ where $X$ is a set, is $\varnothing$ a differentiable function? If so, what is its derivative?

Also, is the empty set a differentiable manifold? If so, what is its dimension?

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    $\begingroup$ In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question. $\endgroup$ Commented Oct 24, 2018 at 23:24
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    $\begingroup$ Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will. $\endgroup$ Commented Oct 24, 2018 at 23:25
  • $\begingroup$ @EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map. $\endgroup$
    – Tyler
    Commented Oct 24, 2018 at 23:27
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    $\begingroup$ The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about. $\endgroup$ Commented Oct 24, 2018 at 23:51
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    $\begingroup$ To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$. $\endgroup$ Commented Oct 25, 2018 at 3:03

2 Answers 2


Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $\emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)

For any smooth manifold $X$, the empty function $\emptyset\to X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $\emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $\emptyset$ as an open subset of $\mathbb{R}^m$, then the empty function $\emptyset\to\mathbb{R}^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $\emptyset\to \mathbb{R}^{n\times m}$.)

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    $\begingroup$ I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $\emptyset'=\emptyset.$ $\endgroup$ Commented Oct 25, 2018 at 4:49

Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.

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    $\begingroup$ Yes. Because there are no points in its domain where it fails to have a derivative. $\endgroup$ Commented Oct 25, 2018 at 4:52
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    $\begingroup$ +1. Does it mean it is also "nowhere differentiable"? $\endgroup$
    – Taladris
    Commented Oct 26, 2018 at 0:33
  • $\begingroup$ @DanielWainfleet With same logic we can assert, that at every point in the empty set, it is not differentiable. With same success every element in the empty set is triangle and is not triangle. Both is true? $\endgroup$
    – zkutch
    Commented Mar 7, 2021 at 20:48

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