# nonempty perfect subset of $\mathbb{R}$ that contains no rational number, a

We will construct a nonempty perfect set^{} contained in $\mathbb{R}$ that
contains no rational number.

We will begin with a closed interval^{}, and then, imitating the
construction of Cantor set, we will inductively delete each
rational number in it together with an open interval. We will do
it in such a way that the end points of the open intervals will
never be deleted afterwards.

Let ${E}_{0}=[{b}_{0},{a}_{0}]$ for some irrational numbers ${a}_{0}$ and ${b}_{0}$, with $$ . Let $\{{q}_{1},{q}_{2},{q}_{3},\mathrm{\dots}\}$ be an enumeration of the rational numbers in $[{b}_{0},{a}_{0}]$. For each ${q}_{i}$, we will define an open interval $({a}_{i},{b}_{i})$ and delete it.

Let ${a}_{1}$ and ${b}_{1}$ be two irrational numbers such that $$. Define ${E}_{1}={E}_{0}\backslash ({a}_{1},{b}_{1})$.

Having defined ${E}_{1},{E}_{2},\mathrm{\dots},{E}_{n}$, ${a}_{1},{a}_{2},\mathrm{\dots},{a}_{n}$ and ${b}_{1},{b}_{2},\mathrm{\dots},{b}_{n}$, let’s define ${a}_{n+1}$ and ${b}_{n+1}$:

If ${q}_{n+1}\in {\displaystyle \bigcup _{k=1}^{n}}({a}_{k},{b}_{k})$ then there exists an $i\le n$ such that ${q}_{n+1}\in ({a}_{i},{b}_{i})$. Let ${a}_{n+1}={a}_{i}$ and ${b}_{n+1}={b}_{i}$.

Otherwise let ${a}_{n+1}$ and ${b}_{n+1}$ be two irrational numbers such that $$, and which satisfy:

$$ |

and

$$ |

Now define ${E}_{n+1}={E}_{n}\backslash ({a}_{n+1},{b}_{n+1})$. Note that by our choice of ${a}_{n+1}$ and ${b}_{n+1}$ any of the previous end points are not removed from ${E}_{n}$.

Let $E={\displaystyle \bigcap _{n=1}^{\mathrm{\infty}}}{E}_{n}$. $E$ is clearly
nonempty, does not contain any rational number, and also it is
compact^{}, being an intersection^{} of compact sets.

Now let us see that $E$ does not have any isolated points. Let
$x\in E$, and $\u03f5>0$ be given. If $x\ne {a}_{j}$ for any $j\in \{0,1,2,\mathrm{\dots}\}$, choose a rational number
${q}_{k}$ such that $$. Then ${q}_{k}\in ({a}_{k},{b}_{k})$ and
since $x\in E$ we must have $$, which means ${a}_{k}\in (x,x+\u03f5)$. Since we know that ${a}_{k}\in E$, this shows that $x$ is a limit point^{}. Otherwise, if $x={a}_{j}$ for some $j$, then choose a ${q}_{k}$ such that $$. Similarly, ${q}_{k}\in ({a}_{k},{b}_{k})$ and it follows that ${b}_{k}\in (x-\u03f5,x)$. We have shown that
any point of $E$ is a limit point, hence $E$ is perfect.

Title | nonempty perfect subset of $\mathbb{R}$ that contains no rational number, a |
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Canonical name | NonemptyPerfectSubsetOfmathbbRThatContainsNoRationalNumberA |

Date of creation | 2013-03-22 15:26:10 |

Last modified on | 2013-03-22 15:26:10 |

Owner | Gorkem (3644) |

Last modified by | Gorkem (3644) |

Numerical id | 18 |

Author | Gorkem (3644) |

Entry type | Example |

Classification | msc 54A99 |