Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there a $p$-adic analogue to the intermediate value theorem? I know there is a notion of convex sets in the $p$-adic context but can we hope for an intermediate value theorem in this context?

share|cite|improve this question
Could you explain to the uninformed like me what a $p$-adic convex set is? Also: isn't the proof of the intermediate value theorem an argument that ultimately boils down to connectivity? Since $p$-adic things tend to be totally disconnected, I'm wondering what kind of statement you're looking for. – t.b. Nov 5 '11 at 16:52
Let $a$ and $b$ be in $\mathbb{Q}_p$. Then an interval $[a,b]$ is the smallest ball containing both $a$ and $b$. A convex set is a set which for every points $a,b$ in it $[a,b]$ is contained in the set. I am looking for a statement like: if $A \subset \mathbb{Q}_p$ is convex, then $f(A)$ is convex. Clearly this is false in general, but if $f$ is $C^1$ then for sufficiently small balls $A$ $f(A)$ is also a ball, so it is convex. – user17090 Nov 5 '11 at 17:08

This is not true even for polynomial functions. Consider $A=\mathbb Z_p=[0,1]$ the unit closed disk and $f(z)=z^2$. Then $f(0)=0, f(1)=1$ but $[0,1]$ is not contained in $f(A)$: $p\in [0,1]\setminus f(A)$.

edit It is not even true that $f(A)$ is a ball for small enough ball $A$. In the above example, take $A=[0,b]$ for any non zero $b$, let $n$ be an odd natural integer such that $n>v_p(b^2)$, then $p^n \in [0, f(b)] \setminus f(A)$.

You will have better chance if you work with $\mathbb C_p$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.