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# Dense subset of R

A set $$A$$ is said to be a dense subset of $$R$$ if $$\forall x,y\in R,x<y$$, $$\exists a\in A$$, such that $$x<a<y$$. My this note is to show that the set $$S=\left\{ m+n\sqrt { 2 } :m,n\in Z \right\}$$ is dense in $$R$$. For that we observe that $$S$$ satisfies the following properties:

i)Additivity: If $$x,y\in S$$, then $$\left( x+y \right) \in S$$.

ii)Homogeneity in $$Z$$: If $$x\in S$$ and $$k\in Z$$, then $$kx\in S$$

Now we observe that $$\forall n\in Z$$, $$0\le \left( n\sqrt { 2 } -\left\lfloor n\sqrt { 2 } \right\rfloor \right) < 1$$.

Also we note that $$\forall m,n,m',n'\in Z$$, $m+n\sqrt { 2 } =m'+n'\sqrt { 2 } \\ \Rightarrow m=m'\quad and\quad n=n'$

Now, let $${ s }_{ n }=\left( n\sqrt { 2 } -\left\lfloor n\sqrt { 2 } \right\rfloor \right) ,n\in Z$$ Then $$0\le { s }_{ n }< 1$$ and $${ s }_{ n }\in S$$. Also if $$n\neq n'$$, then $${ s }_{ n }\neq { s }_{ n' }$$.Hence $$\left\{ { s }_{ n }:n\in Z \right\}$$ is an infinite subset of $$S\cap [0,1)$$.

Now we shall show that for any $$\varepsilon >0$$, $$\exists s\in S$$ such that $$0<s<\varepsilon$$. For that, given any $$\varepsilon >0$$, we partition $$[0,1)$$ into $$k$$ equal subintervals, each having size less than $$\varepsilon$$. Then, since $$\left\{ { s }_{ n }:n\in Z \right\}$$ is an infinite subset of $$S\cap [0,1)$$, so at least one subinterval must contain two distinct elements of $$S\cap [0,1)$$, say $${ s }_{ n },{ s }_{ n' }$$. Without loss of generality, let $${ s }_{ n }<{ s }_{ n' }$$. Then obviously $$0<{ s }_{ n' }-{ s }_{ n }<\varepsilon$$. Now, from the aforesaid properties i) and ii) of $$S$$, we have that $$\left( { s }_{ n' }-{ s }_{ n } \right) =s\in S$$ and hence $$0<s<\varepsilon$$. This shows that for any $$\varepsilon >0$$, $$\exists s\in S$$ such that $$0<s<\varepsilon$$.

Now let $$\varepsilon >0$$ be a real to be chosen such that $$b-a>\varepsilon ,a,b\in R$$ be given. Then it is trivial to show that $$\exists {n}_{1}\in Z$$ such that $$a<{n}_{1}\varepsilon <b$$. Now we take $$\varepsilon =\frac { b-a }{ 2 }$$. Then $$\exists s\in S$$ such that $$0<s<\varepsilon$$. Hence, $$\exists {n}_{2}\in Z$$ such that $$a<{n}_{2}s<b$$. Since by property ii), $${n}_{2}s\in S$$,the property is proved.

There is another way to prove the above proposition. First prove that every proper subgroup of the additive group $$R$$ is either dense or cyclic in $$R$$. Then prove that the given subset $$S$$ of $$R$$ is a subgroup of $$R$$ that is not cyclic in $$R$$..hence it is dense in $$R$$.

Note by Kuldeep Guha Mazumder
2 years, 2 months ago

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