Dense subset of R

A set AA is said to be a dense subset of RR if x,yR,x<y\forall x,y\in R,x<y, aA\exists a\in A, such that x<a<yx<a<y. My this note is to show that the set S={m+n2:m,nZ}S=\left\{ m+n\sqrt { 2 } :m,n\in Z \right\} is dense in RR. For that we observe that SS satisfies the following properties:

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

ii)Homogeneity in ZZ: If xSx\in S and kZk\in Z, then kxSkx\in S

Now we observe that nZ\forall n\in Z, 0(n2n2)<10\le \left( n\sqrt { 2 } -\left\lfloor n\sqrt { 2 } \right\rfloor \right) < 1.

Also we note that m,n,m,nZ\forall m,n,m',n'\in Z, m+n2=m+n2m=mandn=nm+n\sqrt { 2 } =m'+n'\sqrt { 2 } \\ \Rightarrow m=m'\quad and\quad n=n'

Now, let sn=(n2n2),nZ{ s }_{ n }=\left( n\sqrt { 2 } -\left\lfloor n\sqrt { 2 } \right\rfloor \right) ,n\in Z Then 0sn<10\le { s }_{ n }< 1 and snS{ s }_{ n }\in S. Also if nnn\neq n', then snsn{ s }_{ n }\neq { s }_{ n' }.Hence {sn:nZ}\left\{ { s }_{ n }:n\in Z \right\} is an infinite subset of S[0,1)S\cap [0,1) .

Now we shall show that for any ε>0\varepsilon >0, sS\exists s\in S such that 0<s<ε0<s<\varepsilon . For that, given any ε>0\varepsilon >0, we partition [0,1)[0,1) into kk equal subintervals, each having size less than ε\varepsilon . Then, since {sn:nZ}\left\{ { s }_{ n }:n\in Z \right\} is an infinite subset of S[0,1)S\cap [0,1) , so at least one subinterval must contain two distinct elements of S[0,1)S\cap [0,1) , say sn,sn{ s }_{ n },{ s }_{ n' }. Without loss of generality, let sn<sn{ s }_{ n }<{ s }_{ n' }. Then obviously 0<snsn<ε0<{ s }_{ n' }-{ s }_{ n }<\varepsilon . Now, from the aforesaid properties i) and ii) of SS, we have that (snsn)=sS\left( { s }_{ n' }-{ s }_{ n } \right) =s\in S and hence 0<s<ε0<s<\varepsilon . This shows that for any ε>0\varepsilon >0, sS\exists s\in S such that 0<s<ε0<s<\varepsilon .

Now let ε>0\varepsilon >0 be a real to be chosen such that ba>ε,a,bRb-a>\varepsilon ,a,b\in R be given. Then it is trivial to show that n1Z\exists {n}_{1}\in Z such that a<n1ε<ba<{n}_{1}\varepsilon <b. Now we take ε=ba2\varepsilon =\frac { b-a }{ 2 } . Then sS\exists s\in S such that 0<s<ε0<s<\varepsilon . Hence, n2Z\exists {n}_{2}\in Z such that a<n2s<ba<{n}_{2}s<b. Since by property ii), n2sS{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 RR is either dense or cyclic in RR. Then prove that the given subset SS of RR is a subgroup of RR that is not cyclic in RR..hence it is dense in RR.

Note by Kuldeep Guha Mazumder
4 years, 5 months ago

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