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
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 $\bar E$.
1. 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
$$\bex E=\sex{\bigcup_{n\geq 1}(a_n,b_n)}^c, \eex$$
where $(a_i,b_i)$, $(a_j,b_j)$ ($i\neq j$) have no common points.
2. 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,
$$\bex E'^c=\bigcup_{n\geq 1}(a_n,b_n). \eex$$
For $\forall\ x\in E'$, $\forall\ \delta>0$, we have
$$\bee\label{dec} [x-\delta,x+\delta]\cap E'=\sex{[x-\delta,x+\delta]\cap (E'\bs E)} \cup\sex{[x-\delta,x+\delta]\cap E}. \eee$$
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
$$\bex [x-\delta,x+\delta]\cap (E'\bs E)\neq \vno. \eex$$
This completes the proof of Theorem 1.
