# Submmersions: $|f(x)|$ do not assume maximal value

I am studying analysis and I have had a lot of uncertainties. For instance, I cannot solve this exercise:

If $$f:U\rightarrow\mathbb{R}^3$$ has class $$C^1$$ and rank $$3$$ in all of the points of the open $$U\in\mathbb{R^4}$$, show that $$|f(x)|$$ do not assume maximal value for $$x\in U$$.

(I guess this is the comand, but I'm so sorry if I did mistakes. My language and the language of the comand is Portuguese)

Well. I know that $$f$$ is a submersion. So, it's an open map. From here can I get the required? If I know that $$f$$ is an open map, have I that $$|f(x)|$$ is an open set and so that it has not a maximum?

Edit - September, 25

I was taking another look at this question I decided try a formal proof in here:

Once the rank of $$f$$ is maximal (the dominium dimension) and the contradominium has a lower dimension, so $$f$$ is a submmersion. But the submmersions are opened applications. So, $$A=\{f(x);x\in U\}\subset \mathbb{R}^4$$ is open.

Consider a value $$|f(x_0)|,x_0\in U$$ and we'll prove that it isn't maximal. Once $$A$$ is open, there is $$\delta>0$$ such that $$B[f(x_0),\delta]\subset A$$.

Taking $$f(x_0)$$ as a vector, we have $$(1+\delta)f(x_0)\in A$$. So, there is $$x_1\in U$$ such that $$f(x_1)=(1+\delta)f(x_0)$$.

But $$|f(x_1)|=(1+\delta)|f(x_0)|>|f(x_0)|$$ and this completes the proof.

What do you think? Thanks very much.

• Your second paragraph makes no sense to me. – zhw. Jul 16 '18 at 20:08
• Sorry, I did a mistake and I hope now the question is correct. Thanks. – Na'omi Jul 16 '18 at 21:52
• Your proof is correct. The idea is very geometric, once you know how to prove that submersions are open maps. – Laz Sep 25 '18 at 22:06
• @Laz, thanks very much for the reply. – Na'omi Sep 25 '18 at 22:51