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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
1 year, 9 months ago

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