Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm wondering if the set of all rationals with denominators less than $10^6$ is closed in the real number system. I think it is not, so that's what I've been trying to prove. I've tried looking at its complement and showing that it is open, but I didn't progress much. My last resort is to show that there is a sequence of rationals in my set converging to an irrational number (or to a rational whose denominator is greater than $10^6$). The problem is, how do I ensure that all simplified fractions in my sequence will have denominator less than $10^6$?

share|improve this question
add comment

5 Answers

up vote 16 down vote accepted

Hint: Let $N=\operatorname{lcm}\{1,2,3,\ldots,10^6\}$, and let our set be $$S=\{\tfrac{a}{b}\in\mathbb{Q}\mid 1\leq b\leq 10^6\}.$$

Consider the homeomorphism $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=Nx$. Then $f(S)\subseteq\mathbb{Z}$.

share|improve this answer
    
+1 that's wonderful! –  user12014 Jul 6 '12 at 22:17
    
that's pretty smooth. thanx –  The Substitute Jul 6 '12 at 22:30
add comment

Of course it's closed. Just blow it up by a factor of $(10^6)!$, and all its elements are integers.

share|improve this answer
    
Nice, I missed the (on retrospect) clearest reason. –  André Nicolas Jul 6 '12 at 23:44
add comment

Call our set $D$. We show that $D$ is closed by showing that the closure of $D$ is equal to $D$.

Any bounded set $B$ contains only finitely many elements of $D$. For suppose to the contrary that $B$ contains infinitely many elements of $D$. Since the number of denominators is finite, the number of numerators of elements of $B$ must be infinite, and therefore unbounded, making $B$ unbounded.

So if we take any real number $x$ not in $D$, there is an open interval about $x$ which does not contain any element of $D$. Thus $x$ cannot be in the closure of $D$.

share|improve this answer
    
Aren't you really showing the complement is open directly? –  Arturo Magidin Jul 6 '12 at 22:10
    
@ Andre; by finite intervals, do you mean bounded? And how do you know there will not be infinitely many rationals with denominators less than $10^6$ in that interval? –  The Substitute Jul 6 '12 at 22:11
    
Yes. I wanted to write along lines the OP had explored. –  André Nicolas Jul 6 '12 at 22:11
    
If the set of denominators is finite, and our set of fractions is infinite, then there must be infinitely many numerators, making the set unbounded. –  André Nicolas Jul 6 '12 at 22:14
    
@Broseph: Between $0$ and $1$, there are only finitely many such rationals. Therefore, between $n$ and $n+1$ (for $n$ an integer) there are only finitely many such rationals. Therefore, between $-n$ and $n$ ($n$ an integer), there are only fintiely many such rationals. –  Arturo Magidin Jul 6 '12 at 22:15
add comment

Note that $D= \displaystyle \bigcup_{n=1}^{10^6} \frac{1}{n}\mathbb{Z}$. Each element of this union is closed since $\mathbb{Z}$ is closed and the map $x \mapsto \frac{x}{n}$ is a homeomorphism for all $n \ne 0$, and a finite union of closed sets is closed.

share|improve this answer
add comment

Lemma: Suppose $X,Y\subseteq \mathbb R$, $X$ is finite and $Y$ is closed. Then the set $$X+Y:=\{x+y|x\in X, y\in Y\}$$ is closed.

Proof: Since $Y$ is closed, any sequence $(y_n)$ in $Y$ with limit in $\mathbb R$ has limit in $(y_n)$. For $x\in X$, any sequence in $\{x\}+Y$ is of the form $(x+y_n)$, and if $(x+y_n)$ has limit $x+y$ in $\mathbb R$ then $(y_n)$ has limit $y\in Y$, so $x+y\in \{x\}+Y$ hence $(x+y_n)$ has limit in $\{x\}+Y$. Thus $\{x\}+Y$ is closed. Since a finite union of closed sets is closed, it follows that $X+Y=\bigcup\limits_{x\in X} \{x\}+Y$ is closed.

Corollary: The set $\{a/b|a,b\in \mathbb Z, 0\leq b\leq 10^6\}$ is closed.

Proof: Note that $$\{a/b|a,b\in \mathbb Z, 0\leq b\leq 10^6\}=\{a/b|a,b\in \mathbb Z, 0\leq a\leq b\leq 10^6\}+\mathbb Z$$ and $\{a/b|a,b\in \mathbb Z, 0\leq a\leq b\leq 10^6\}$ is finite while $\mathbb Z$ is closed.

share|improve this answer
add comment

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.