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,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
5 years, 7 months ago

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Your proof is very good in all respect. But I think there is some error at the very last paragraph. You have taken ε=ba2\varepsilon = \frac {b-a} {2} so that 0<ε<ba.0 \lt \varepsilon \lt b - a. Then there exists n2Zn_2 \in \Bbb Z such that a<n2ε<b.a \lt n_2 \varepsilon \lt b. Now since (0,ε)S(0, \varepsilon) \cap S \neq \varnothing there exists sSs \in S such that 0<s<ε.0 \lt s \lt \varepsilon. Now suppose n20n_2 \geq 0 then clearly n2sn2ε<b.n_2 s \leq n_2 \varepsilon < b. How do I ascertain that n2s>a n_2 s \gt a\ ? For instance we could take a=13a = \frac 1 3 and b=12.b = \frac 1 2. Then it is easy to check that n2=5.n_2 = 5. According to your claim we get sSs \in S such that 0<s<112.0 \lt s \lt \frac {1} {12}. Then we have n2s=5s<512<12.n_2 s = 5 s \lt \frac {5} {12} \lt \frac {1} {2}. But how do I ascertain that 5s>13.5 s \gt \frac 1 3. For that we need s>115.s \gt \frac {1} {15}. How do you know that there is such an sSs \in S with 115<s<112 \frac {1} {15} \lt s \lt \frac {1} {12}\ ? Thus in all possible cases how do I ascertain that there exists such an sSs \in S such that n2s>a.n_2 s \gt a. Also if n2<0n_2 \lt 0 then multiplying ss by n2n_2 we have n2s>n2ε.n_2 s \gt n_2 \varepsilon. We have to deal with this fact too.

Hoping to get your response in this regard as soon as possible. Thanks in advance.

Arnab Chatterjee - 10 months ago

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