# limit points of uncountable subsets of R^n

Proposition^{}. Let ${\mathbb{R}}^{n}$ be an $n$-dimensional, real normed space and let $A\subseteq {\mathbb{R}}^{n}$. If $A$ is uncountable, then there exists limit point^{} of $A$ in ${\mathbb{R}}^{n}$.

Proof. For any $k\in \mathbb{N}$ let

$${\mathbb{B}}_{k}=\{v\in {\mathbb{R}}^{n}|||v||\le k\},$$ |

i.e. ${\mathbb{B}}_{k}$ is a closed ball^{} centered in $0$ with radius $k$. Assume, that for any $k$ the set

$${V}_{k}={\mathbb{B}}_{k}\cap A$$ |

is finite. Then $\bigcup {V}_{k}=A$ would be at most countable^{}. Contradiction^{}, since $A$ is uncountable. Thus, there exists ${k}_{0}\in \mathbb{N}$ such that ${V}_{{k}_{0}}$ is infinite^{}. But ${V}_{{k}_{0}}\subseteq {\mathbb{B}}_{{k}_{0}}$ and since ${\mathbb{B}}_{{k}_{0}}$ is compact^{} (and ${V}_{{k}_{0}}$ is infinite), then there exists limit point of ${V}_{{k}_{0}}$ in ${\mathbb{R}}^{n}$. This completes^{} the proof. $\mathrm{\square}$

Corollary. If $A\subseteq {\mathbb{R}}^{n}$ is uncountable, then there exist infinitely many limit points of $A$ in ${\mathbb{R}}^{n}$.

Proof. Assume, that there are finitely many limit points of $A$, namely ${x}_{1},\mathrm{\dots},{x}_{k}\in {\mathbb{R}}^{n}$. For $\epsilon >0$ define

$${A}_{\epsilon}=\{v\in {\mathbb{R}}^{n}|{\forall}_{i}||v-{x}_{i}||>\epsilon \}.$$ |

Briefly speaking, ${A}_{\epsilon}$ is a complement^{} of a union of closed balls centered at ${x}_{i}$ with radii $\epsilon $. Of course ${A}_{\epsilon}\ne \mathrm{\varnothing}$ since there are finitely many limit points. Let

$${V}_{\epsilon}=A\cap {A}_{\epsilon}.$$ |

Assume, that ${V}_{\epsilon}$ is countable for every $\epsilon $. Then

$$A\subseteq \bigcup _{n\in \mathbb{N}}{V}_{\frac{1}{n}}\cup \{{x}_{1},\mathrm{\dots},{x}_{k}\}$$ |

would be at most countable (of course under assumption^{} of Axiom of Choice^{}). Contradiction. Thus, there is $\gamma >0$ such that ${V}_{\gamma}$ is uncountable. Then (due to proposition) there is a limit point ${x}^{\prime}\in {\mathbb{R}}^{n}$ of ${V}_{\gamma}$. Note, that

$${x}^{\prime}\in \overline{{V}_{\gamma}}\subseteq {V}_{{\gamma}^{\prime}}$$ |

for some $$. Thus ${x}^{\prime}$ is different from any ${x}_{i}$. Contradiction, since ${x}^{\prime}$ is also a limit point of $A$. $\mathrm{\square}$

Title | limit points of uncountable subsets of R^n |
---|---|

Canonical name | LimitPointsOfUncountableSubsetsOfRn |

Date of creation | 2013-03-22 19:07:57 |

Last modified on | 2013-03-22 19:07:57 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 6 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54A99 |