# Why do we want TWO open sets from the inverse function theorem?

I have been analyzing Rudin's proof of the Inverse Function Theorem closely over the last two days, and trying to understand what the purpose of every assumption made is.

The first assumption that he makes is the value of the radius $\lambda$ of the neighborhood U such that $2\lambda = \frac{1}{\lVert f'(a)^{-1} \rVert}$.

This definition of lambda helps in large part because it ensures that $f'$ is invertible for every point in the open neighborhood selected, using the fact that the space of invertible linear operators is open, as well as the specific bound (Rudin Theorem 9,8):

(1) $\lVert f'(x) - f'(a)\rVert < \frac{1}{\lVert f'(a)^{-1} \rVert}$.

With the choice of open neighborhood given in Rudin (which is possible only due to the continuous differentiability of f at a, since normally the ball of a pseudometric need not be open) we get for all x in the chosen open neighborhood:

(2) $\lVert f'(x) - f'(a)\rVert \le (\frac{1}{2}) \frac{1}{\lVert f'(a)^{-1} \rVert}$

Which clearly satisfies the given requirement, although it is obviously stronger. So I was looking for the other conclusions of the theorem that required this stronger condition, and it seems to me that it is only needed for showing that the image of the open neighborhood f(U)=V is open.

Now don't get me wrong, I get why this is a neat and interesting property for f(U) to have. But we don't need it for f to be invertible. We don't need it for its inverse to be differentiable, and hence for f to be a homeomorphism (i.e. for the inverse to be continuous as a function restricted to the image). We don't need it for $f'$ to be a homeomorphism (only that f is continuously differentiable on the entire neighborhood U).

So why ask for it at all? All it does, as far as I can see, is obscure the important connection between the proof of the Inverse function theorem and the openness of the space of linear operators.

I.e., any lambda such that $c\lambda = \frac{1}{\lVert f'(a)^{-1} \rVert}$ for c>1 satisfies (1), which is all that is necessary for all of the important conclusions of the theorem. Whereas (2) holds only for $c \ge 2$.

So any way, why should we want f(U) to be open so badly? Why not just satisfy (1)?

• The standard phrasing of the Inverse Function Theorem is that if $f'(a)$ is invertible, then there are open sets $U$ around $a$ and $V$ around $f(a)$ such that the restricted map $f|_U \colon U \to V$ is a diffeomorphism. Does that phrasing clarify things? Apr 9, 2016 at 2:06
• No not at all. The point of the question is that we can relax the assumptions and substantially simplify the proof by not requiring that V be open. f restricted to U would even still be a diffeomorphism (I think). So why care so much about whether V is open or not? Apr 9, 2016 at 14:41
• Maybe this is overkill, and it doesn't quite answer your question, but the following theorem is relevant: en.wikipedia.org/wiki/Invariance_of_domain In particular: Let $U \subset \mathbb{R}^n$ be open. If $f \colon U \to f(U)$ is a homeomorphism, then $f(U)$ is automatically open. Contrapositive: if $f(U)$ is not open, then $f \colon U \to f(U)$ is not a homeomorphism. Apr 9, 2016 at 22:25
• Looking at the definition of limit for a metric space, the neighborhoods have to approach the limit point in the co-domain space, not just the range space, so using that definition $f^{-1}$ is definitely neither differentiable nor continuous. This also explains why Rudin proves openness of f(U) before differentiability of $f^{-1}$. And even redefining f so that the codomain is f(U) and using the subspace metric on f(U) etc., whatever we could only get "differentiability" of $f^{-1}$, not continuity (or else we would be contradicting invariance of domain theorem) and even this differentiability Apr 9, 2016 at 22:59
• restricted to f(U) would not be at all interesting or useful since, f(U) would no longer be homeomorphic to any R^n or open subset thereof, and hence would be of no use, for example, in analyzing manifolds. This helps a lot; thank you so much for your help @JesseMadnick! Apr 9, 2016 at 23:02

We do actually use the openness of $V$ when proving $f$ has a differentiable inverse, just that it's never explicitly stated when used. By definition of differentiability (using the notation in the book), we need to show in part (b) that for every $y \in V,$

$$\frac{|g(y+k)-g(y)-Tk|}{|k|} \rightarrow 0$$

as $k \rightarrow 0.$ For this to make sense however, we require $y+k \in V$ for sufficiently small $k$ for $g$ to be defined (recall the $\varepsilon-\delta$ definition of a limit if this isn't clear). But this is exactly the same as requiring $f(V)$ to be open.

• I agree with you that if one takes Rudin's definition of differentiability (p.212 Definition 9.11) as gospel, one requires even for point-wise differentiability that f(a) is in an open set, a fact which I had not previously considered. So even if V not being open is fine for f being differentiable, we need it for g to be differentiable using Rudin's definition. But why can't use a more general definition? (See for example: math.stackexchange.com/questions/161910/…) The link says that the neighborhoods would not be Apr 9, 2016 at 16:42
• homeomorphic to open intervals/sets, but I don't see why not, since seemingly f and g=$f^{-1}$ would both be continuous on U and f(U) even using the more general definition of differentiability for the derivative of g (although continuity on f(U) would seemingly be a weaker condition than continuity on an open set, since there would be no possible continuous extension to all of R^n in general). Apr 9, 2016 at 16:44
• That is true, by redefining the limit it seems like you can allow $\lambda$ to be larger. It seems like functions that are diffeomorphic in this sense aren't quite as well-behaved though, which is worth keeping in mind.
– ktoi
Apr 9, 2016 at 17:14
• And we need to have two open sets to define coordinate/charts/local coordinates for manifolds? I'm trying to begin seriously learning manifold theory, hence why I am so insistent upon understanding the inverse function theorem well. Apr 9, 2016 at 17:50

It seems to me that $c=\frac{1}{2}$ is essential for then proving line number (51). In the line before, note Rudin write "By (50), $\|h - A^{-1}k\| \leq \frac{1}{2} \|h\|$. That factor of 1/2 comes exactly from the c=2 you were mentioning earlier. And it's only having this factor of 1/2 that Rudin can then conclude $\|A^{-1} k\| \geq \frac{1}{2} \|h\|$, and get the nice (51).

• I thought so too the first time I looked at the proof. However, when I looked at it more closely, it seems like we just need $|A^{-1}k| \ge c^{-1}$ in order to derive (51) $|h| \le c \lVert A^{-1} \rVert |k| = \lambda^{-1} |k|$. and the rest of the proof seems to go through as long as $\lambda^{-1} > 1$ or equivalently $c>1$. Apr 9, 2016 at 16:46