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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 $10^6$?

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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 $f(S)\subseteq\mathbb{Z}$.

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  • $\begingroup$ that's pretty smooth. thanx $\endgroup$ – The Substitute Jul 6 '12 at 22:30
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Of course it's closed. Just blow it up by a factor of $(10^6)!$, and all its elements are integers.

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  • $\begingroup$ Nice, I missed the (on retrospect) clearest reason. $\endgroup$ – André Nicolas Jul 6 '12 at 23:44
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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 $D$.

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  • $\begingroup$ Aren't you really showing the complement is open directly? $\endgroup$ – Arturo Magidin Jul 6 '12 at 22:10
  • $\begingroup$ @ 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? $\endgroup$ – The Substitute Jul 6 '12 at 22:11
  • $\begingroup$ Yes. I wanted to write along lines the OP had explored. $\endgroup$ – André Nicolas Jul 6 '12 at 22:11
  • $\begingroup$ 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. $\endgroup$ – André Nicolas Jul 6 '12 at 22:14
  • $\begingroup$ @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. $\endgroup$ – Arturo Magidin Jul 6 '12 at 22:15
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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.

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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.

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