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

The question is how we construct a function $f:\mathbb Q_p\to\mathbb R$ so that $f$ is discontinuous at every $x_0\in\mathbb Q_p$.

share|cite|improve this question
Where does such a question come from? Is it idle speculation? – KCd Nov 17 '11 at 4:21
for each $x \in Q_p$ pick a number $f(x) \in \mathbb R$ in a "random" way... – N. S. Nov 17 '11 at 5:10
How do we pick $f(x)$ in a "random way"? – Jie Fan Nov 17 '11 at 5:12
@KCd: It's easy to contruct a discontinous function from $\mathbb Q_p$ to $\mathbb Q_p$, it's hard from $\mathbb R\to\mathbb R$. The left may be interesting? – Jie Fan Nov 17 '11 at 5:15
You should include such background in the question itself, so the point is clearer. Otherwise it seems like a, well, random question. – KCd Nov 17 '11 at 5:44
up vote 2 down vote accepted

Let $D=\{0,1\}$ with the discrete topology, so that $D^\omega$ with the product topology is a Cantor set. $\mathbb{Q}_p$ is homeomorphic to $\omega\times D^\omega$ or, equivalently, to $D^\omega\setminus\{p\}$ for any $p\in D^\omega$. In particular, it has a countable dense subset $S$, and $\mathbb{Q}_p\setminus S$ is also dense in $\mathbb{Q}_p$. In fact it’s well-known that $\mathbb{Q}$ is dense in $\mathbb{Q}_p$, so we may take $S=\mathbb{Q}$, but any countable dense $S$ will work equally well.

Then $$\chi_S:\mathbb{Q}_p\to\mathbb{R}:x\mapsto\begin{cases}1,&x\in S\\ 0,&x\notin S\;, \end{cases}$$

the indicator (characteristic) function of $S$, is nowhere continuous, and in particular $\chi_\mathbb{Q}$ is nowhere continuous.

share|cite|improve this answer

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.