limit points of uncountable subsets of R^n

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

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

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. $\square$

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 $x_{1},\ldots,x_{k}\in\mathbb{R}^{n}$. For $\varepsilon>0$ define

 $A_{\varepsilon}=\{v\in\mathbb{R}^{n}\ |\ \forall_{i}\ ||v-x_{i}||>\varepsilon\}.$

Briefly speaking, $A_{\varepsilon}$ is a complement of a union of closed balls centered at $x_{i}$ with radii $\varepsilon$. Of course $A_{\varepsilon}\neq\emptyset$ since there are finitely many limit points. Let

 $V_{\varepsilon}=A\cap A_{\varepsilon}.$

Assume, that $V_{\varepsilon}$ is countable for every $\varepsilon$. Then

 $A\subseteq\bigcup_{n\in\mathbb{N}}V_{\frac{1}{n}}\cup\{x_{1},\ldots,x_{k}\}$

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 $0<\gamma^{\prime}<\gamma$. Thus $x^{\prime}$ is different from any $x_{i}$. Contradiction, since $x^{\prime}$ is also a limit point of $A$. $\square$

Title limit points of uncountable subsets of R^n LimitPointsOfUncountableSubsetsOfRn 2013-03-22 19:07:57 2013-03-22 19:07:57 joking (16130) joking (16130) 6 joking (16130) Theorem msc 54A99