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


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

     

    Zujin Zhang

     

    School of Mathematics and Computer Science,

    Gannan Normal University

    Ganzhou 341000, P.R. China

     

    zhangzujin361@163.com

     

    MSC2010: 26A03.

     

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

     

    Abstract:

    Let $Esubset bR^1$ be non-empty, countable, dense-in-self, then we shall show that $ar Es E$ is dense in $ar E$.

     

    1. Introduction and the main result

     

     As is well-known, $bQsubsetbR^1$ is countable, dense-in-self (that is, $bQsubset bQ'=bR^1$); and $bR^1s bQ$ is dense in $bR^1$.

     

     We generalize this fact as

    Theorem 1. Let $Esubset bR^1$ be non-empty, countable, dense-in-self, then $ar Es E$ is dense in $ar E$.

     

    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 $Esubset E'$; 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 $bR^1$.

     

    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

     $$ex E=sex{igcup_{ngeq 1}(a_n,b_n)}^c, eex$$

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

     

    2. Proof of Theorem 1。

    Since $E$ is dense-in-self, we have $Esubset E'$, $ar E=E'$. Also, by the fact that $E''=E'$, we see $E'$ is complete, and has power $c$. Note that $E$ is countable, we deduce $E's E eq vno$.

     

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

    $$ex E'^c=igcup_{ngeq 1}(a_n,b_n). eex$$

    For $forall xin E'$, $forall delta>0$, we have

     $$eelabel{dec} [x-delta,x+delta]cap E'=sex{[x-delta,x+delta]cap (E's E)} cupsex{[x-delta,x+delta]cap E}. eee$$

    By analyzing the complement of $[x-delta,x+delta]cap (E's E)$, 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

     $$ex [x-delta,x+delta]cap (E's E) eq vno. eex$$

    This completes the proof of Theorem 1.

     

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  • 原文地址:https://www.cnblogs.com/zhangzujin/p/3703712.html
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