# Dense subset of R

A set $A$ is said to be a dense subset of $R$ if $\forall x,y\in R,x, $\exists a\in A$, such that $x. 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. 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. This shows that for any $\varepsilon >0$, $\exists s\in S$ such that $0.

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 . Now we take $\varepsilon =\frac { b-a }{ 2 }$. Then $\exists s\in S$ such that $0. Hence, $\exists {n}_{2}\in Z$ such that $a<{n}_{2}s. 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
4 years, 2 months ago

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