Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $p:\mathbb{R}^n\to[0,1)^n$ be the map defined by $p(x_1,\ldots,x_n)=(\{x_1\},\ldots,\{x_n\})$, where $\{\cdot\}$ is the fractional part operator. Experimentation suggests that if $S \subseteq \mathbb{R}^n$ is semialgebraic, then the closure of the image set $p(S)$ is semialgebraic. (By a "semialgebraic" subset of $\mathbb{R}^n$ I mean here a finite union of sets, each defined by a system of polynomial inequalities with real coefficients. By "closure" I mean in the usual topology on $\mathbb{R}^n$, restricted to $[0,1)^n$.)

Is this true? Can anyone provide a proof or a counterexample or a relevant reference?

share|improve this question
    
Would S = {(x,y) in R^2 | xy=1} be a counterexample? –  André Henriques Feb 3 '12 at 10:16
    
@André Henriques: Why don't you post that as an answer? It certainly answers the question. –  Ketil Tveiten Feb 3 '12 at 11:06
    
@Andre - Thanks. Looks like I should have done more experimentation! Post it as answer and I'll accept. –  SJR Feb 3 '12 at 11:42

1 Answer 1

up vote 3 down vote accepted

For $n=2$, the set $S=\{(x,y)\in \mathbb R^2| xy=1\}$ provides a counterexample.

For $n=1$, the statement is true.

share|improve this answer

Your Answer

 
discard

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.