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The conclusion of the implicit function theorem for $\mathbb{R}^n$ says that if $F(a, b) = 0$, then there is a function $g$ from an open $V \subseteq \mathbb{R}^m$ to an open $U \subseteq \mathbb{R}^k$ such that $g(b) = a$, and $F(u, v) = 0$ for $u \in U,\ v \in V$ if and only if $g(v) = u$. I've seen one or two source claim that both $U$ and $V$ can be chosen to be open balls. Is this true? Why? Obviously one or the other can be restricted, but how can both be chosen as open balls simultaneously?

Scott M
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3 Answers3

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In general they can't.

The following would be a simple example except that we can take the open sets to be $\mathbb{R}^2$, however it illustrates the issue: $F(x,y) = (x_1-y_1,2x_2-y_2)$.

To fix this, take $F(x,y) = (x_1-y_1, x_2-y_1^2)$, and use the $\|\cdot\|_2$ norm for $x$ and $y$. It is clear that $\frac{\partial f(x,y)}{\partial x} = I $ for all $x,y$, and we must have $g(y) = (y_1, y_2^2)$.

If we take $\hat{y} = (\hat{y}_1 , \hat{y}_2)$ and $V=B(\hat{y}, \epsilon)$, then it is clear that $g(V)$ is not of the form $B(\hat{x}, \eta)$ for some $\hat{x},\eta>0$.

copper.hat
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  • Would it be possible if the (determinant) of the Jacobian is constant? This gives a measure of stretching; so that a ball would be stretched into a ball? – user99680 Dec 19 '13 at 01:45
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    Yes, I believe this is correct, that we can't guarantee that $g(V)$ is a ball. However, what if we allow the codomain to be larger than the image? That is, can we choose $U$ to be an open ball containing $g(V)$, then arbitrarily say that $g:V \to U$? It seems like kind of a dumb solution, but maybe that's what my book's author had in mind. Would it work? – Scott M Dec 19 '13 at 01:55
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    I think it is just sloppiness on the author's part. I would stick with open sets. – copper.hat Dec 19 '13 at 02:20
  • I think @ScottM comment is spot on. We don't require $U$ to be the image $g(V)$. The statement (and proof) are in Sternberg & Loomis, p. 230, and the proof looks fine to me. – Max Apr 06 '17 at 09:17
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The definition of open set and open ball are equivalent since $\mathbb R^n$ is a metric space. Then every open set contains an open ball and every open ball is an open set.

user116007
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  • As I said in the post, obviously you can choose an open ball inside either open set. But how do you know that you can choose both the domain and image to be open sets simultaneously? – Scott M Dec 19 '13 at 00:40
  • Because $U,V$ are open sets you can chose $\varepsilon_U, \varepsilon_V$ sufficiently small, such that the balls are contained in $U,V$. – user116007 Dec 19 '13 at 01:02
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    @user116007: then issue is that the continuous , even differentiable image of an open set is not necessarily open. And I think you need to be more precise about the statement: not every open set in $\mathbb R^n$ is an open ball. – user99680 Dec 19 '13 at 01:22
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Both sets can be open balls inside of the respective open sets $U,V$ ; they can both be open balls because the restriction of the map to $U,V$ is a diffeomorphism, by the inverse function theorem, since the Jacobian matrix $Jf$ is invertible.

user99680
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  • But a diffeomorphism doesn't have to map balls to balls, does it? For example, isn't a circle diffeomorphic to an ellipse? – Scott M Dec 19 '13 at 01:34
  • You're right; let me think it through. I think if the Jacobian is constant , then you can, because the Jacobian measures the rescaling, so that a ball would be rescaled by a constant factor then. – user99680 Dec 19 '13 at 01:39