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

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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 $E\subset \bbR^1$ be non-empty, countable, dense-in-self, then we shall show that $\bar E\bs E$ is dense in $\bar E$ .

Introduction and the main result

As is well-known, $\bbQ\subset\bbR^1$ is countable, dense-in-self (that is, $\bbQ\subset \bbQ‘=\bbR^1$ ); and $\bbR^1\bs \bbQ$ is dense in $\bbR^1$ .

We generalize this fact as

Theorem 1. Let $E\subset \bbR^1$ be non-empty, countable, dense-in-self, then $\bar E\bs E$ is dense in $\bar E$ .

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

Definition 2. A set $E$ is closed iff $E‘\subset E$ . A set $E$ is dense-in-self iff $E\subset E‘$ ; that is, $E$ has no isolated points. A set $E$ is complete iff $E‘=E$ .

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 $E$ has power $c$ ; that is, there is a bijection between $E$ and $\bbR^1$ .

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 $E$ has the form

E = ? ? ? n \geq 1 (a n, b n) ? ? c,

$\bex E=\sex{\bigcup_{n\geq 1}(a_n,b_n)}^c, \eex$

where $(a_i,b_i)$ , $(a_j,b_j)$ ( $i\neq j$ ) have no common points.

Proof of Theorem 1

Since $E$ is dense-in-self, we have $E\subset E‘$ , $\bar E=E‘$ . Also, by the fact that $E‘‘=E‘$ , we see $E‘$ is complete, and has power $c$ . Note that $E$ is countable, we deduce $E‘\bs E\neq \vno$ .

Now that $E‘$ is complete, we see by Lemma 4,

E' c = ? n \geq 1 (a n, b n) .

$\bex E‘^c=\bigcup_{n\geq 1}(a_n,b_n). \eex$

For $\forall\ x\in E‘$ , $\forall\ \delta>0$ , we have

[x ? δ, x + δ] \cap E' = ([x ? δ, x + δ] \cap (E' ? E)) \cup ([x ? δ, x + δ] \cap E) . (1)

$\bee\label{dec} [x-\delta,x+\delta]\cap E‘=\sex{[x-\delta,x+\delta]\cap (E‘\bs E)} \cup\sex{[x-\delta,x+\delta]\cap E}. \eee$

By analyzing the complement of $[x-\delta,x+\delta]\cap (E‘\bs E)$ , we see $[x-\delta,x+\delta]\cap E‘$ (minus $\sed{x-\delta}$ if $x-\delta$ equals some $a_n$ , and minus $\sed{x+\delta}$ if $x+\delta$ equals some $b_n$ ) is compelete, thus has power $c$ . Due to the fact that $E$ is countable, we deduce from $\eqref{dec}$ that

[x ? δ, x + δ] \cap (E' ? E) \neq ? .

$\bex [x-\delta,x+\delta]\cap (E‘\bs E)\neq \vno. \eex$

This completes the proof of Theorem 1.

A fine property of the non-empty countable dense-in-self set in the real line,布布扣,bubuko.com

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

原文：http://www.cnblogs.com/zhangzujin/p/3703712.html

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