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A fine property of the non-empty countable dense-in-self set in the real line

2024-07-02 14:17:05   215人阅读

A fine property of the non-empty countable dense-in-self set in the real line

 

Zujin Zhang

 

School of Mathematics and Computer Science,

GannanNormalUniversity

Ganzhou 341000, P.R. China

 

zhangzujin361@163.com

 

MSC2010: 26A03.

 

Keywords: Dense-in-self set; countable set.

 

Abstract:

Let E?R1<script id="MathJax-Element-1" type="math/tex">E\subset \bbR^1</script> be non-empty, countable, dense-in-self, then we shall show that Eˉ?E<script id="MathJax-Element-2" type="math/tex">\bar E\bs E</script> is dense in Eˉ<script id="MathJax-Element-3" type="math/tex">\bar E</script>.

 

Introduction and the main result

 

 As is well-known, Q?R1<script id="MathJax-Element-4" type="math/tex">\bbQ\subset\bbR^1</script> is countable, dense-in-self (that is, Q?Q′=R1<script id="MathJax-Element-5" type="math/tex">\bbQ\subset \bbQ‘=\bbR^1</script>); and R1?Q<script id="MathJax-Element-6" type="math/tex">\bbR^1\bs \bbQ</script> is dense in R1<script id="MathJax-Element-7" type="math/tex">\bbR^1</script>.

 

 We generalize this fact as

    Theorem 1. Let E?R1<script id="MathJax-Element-8" type="math/tex">E\subset \bbR^1</script> be non-empty, countable, dense-in-self, then Eˉ?E<script id="MathJax-Element-9" type="math/tex">\bar E\bs E</script> is dense in Eˉ<script id="MathJax-Element-10" type="math/tex">\bar E</script>.

 

Before proving Theorem 1, let us recall several related definitions and facts.

 

Definition 2. A set E<script id="MathJax-Element-11" type="math/tex">E</script> is closed iff E′?E<script id="MathJax-Element-12" type="math/tex">E‘\subset E</script>. A set E<script id="MathJax-Element-13" type="math/tex">E</script> is dense-in-self iff E?E′<script id="MathJax-Element-14" type="math/tex">E\subset E‘</script>; that is, E<script id="MathJax-Element-15" type="math/tex">E</script> has no isolated points. A set E<script id="MathJax-Element-16" type="math/tex">E</script> is complete iff E′=E<script id="MathJax-Element-17" type="math/tex">E‘=E</script>.

 

A well-known complete set is the Cantor set. Moreover, we have

 

Lemma 3 ([I.P. Natanson, Theory of functions of a real variable, Rivsed Edition, Translated by L.F. Boron, E. Hewitt, Vol. 1, Frederick Ungar Publishing Co., New York, 1961] P 51, Theorem 1). A non-empty complete set E<script id="MathJax-Element-18" type="math/tex">E</script> has power c<script id="MathJax-Element-19" type="math/tex">c</script>; that is, there is a bijection between E<script id="MathJax-Element-20" type="math/tex">E</script> and R1<script id="MathJax-Element-21" type="math/tex">\bbR^1</script>.

 

 Lemma 4 ([I.P. Natanson, Theory of functions of a real variable, Rivsed Edition, Translated by L.F. Boron, E. Hewitt, Vol. 1, Frederick Ungar Publishing Co., New York, 1961] P 49, Theorem 7). A complete set E<script id="MathJax-Element-22" type="math/tex">E</script> has the form

 

E=???n≥1(an,bn)??c,
<script id="MathJax-Element-23" type="math/tex; mode=display">\bex E=\sex{\bigcup_{n\geq 1}(a_n,b_n)}^c, \eex</script>

where (ai,bi)<script id="MathJax-Element-24" type="math/tex">(a_i,b_i)</script>, (aj,bj)<script id="MathJax-Element-25" type="math/tex">(a_j,b_j)</script> (i≠j<script id="MathJax-Element-26" type="math/tex">i\neq j</script>) have no common points.

 

Proof of Theorem 1

Since E<script id="MathJax-Element-27" type="math/tex">E</script> is dense-in-self, we have E?E′<script id="MathJax-Element-28" type="math/tex">E\subset E‘</script>, Eˉ=E′<script id="MathJax-Element-29" type="math/tex">\bar E=E‘</script>. Also, by the fact that E′′=E′<script id="MathJax-Element-30" type="math/tex">E‘‘=E‘</script>, we see E′<script id="MathJax-Element-31" type="math/tex">E‘</script> is complete, and has power c<script id="MathJax-Element-32" type="math/tex">c</script>. Note that E<script id="MathJax-Element-33" type="math/tex">E</script> is countable, we deduce E′?E≠?<script id="MathJax-Element-34" type="math/tex">E‘\bs E\neq \vno</script>.

 

Now that E′<script id="MathJax-Element-35" type="math/tex">E‘</script> is complete, we see by Lemma 4,

E′c=?n≥1(an,bn).
<script id="MathJax-Element-36" type="math/tex; mode=display">\bex E‘^c=\bigcup_{n\geq 1}(a_n,b_n). \eex</script>

For ? x∈E′<script id="MathJax-Element-37" type="math/tex">\forall\ x\in E‘</script>, ? δ>0<script id="MathJax-Element-38" type="math/tex">\forall\ \delta>0</script>, we have

 

[x?δ,x+δ]∩E′=([x?δ,x+δ]∩(E′?E))∪([x?δ,x+δ]∩E).(1)
<script id="MathJax-Element-39" type="math/tex; mode=display">\bee\label{dec} [x-\delta,x+\delta]\cap E‘=\sex{[x-\delta,x+\delta]\cap (E‘\bs E)} \cup\sex{[x-\delta,x+\delta]\cap E}. \eee</script>

By analyzing the complement of [x?δ,x+δ]∩(E′?E)<script id="MathJax-Element-40" type="math/tex">[x-\delta,x+\delta]\cap (E‘\bs E)</script>, we see [x?δ,x+δ]∩E′<script id="MathJax-Element-41" type="math/tex">[x-\delta,x+\delta]\cap E‘</script> (minus {x?δ}<script id="MathJax-Element-42" type="math/tex">\sed{x-\delta}</script> if x?δ<script id="MathJax-Element-43" type="math/tex">x-\delta</script> equals some an<script id="MathJax-Element-44" type="math/tex">a_n</script>, and minus {x+δ}<script id="MathJax-Element-45" type="math/tex">\sed{x+\delta}</script> if x+δ<script id="MathJax-Element-46" type="math/tex">x+\delta</script> equals some bn<script id="MathJax-Element-47" type="math/tex">b_n</script>) is compelete, thus has power c<script id="MathJax-Element-48" type="math/tex">c</script>. Due to the fact that E<script id="MathJax-Element-49" type="math/tex">E</script> is countable, we deduce from (1)<script id="MathJax-Element-50" type="math/tex">\eqref{dec}</script> that

 

[x?δ,x+δ]∩(E′?E)≠?.
<script id="MathJax-Element-51" type="math/tex; mode=display">\bex [x-\delta,x+\delta]\cap (E‘\bs E)\neq \vno. \eex</script>

This completes the proof of Theorem 1.

 


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