limit points of uncountable subsets of R^n

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

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

$${𝔹}_k=\{v\in {ℝ}^n|||v||\le k\},$$

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

$$V_k={𝔹}_k\cap A$$

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 $V_{k_0}\subseteq {𝔹}_{k_0}$ and since ${𝔹}_{k_0}$ is compact (and $V_{k_0}$ is infinite), then there exists limit point of $V_{k_0}$ in ${ℝ}^n$. This completes the proof. $\mathrm{\square }$

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

Proof. Assume, that there are finitely many limit points of $A$, namely $x_1,\mathrm{\dots },x_k\in {ℝ}^n$. For $\epsilon >0$ define

$$A_{\epsilon }=\{v\in {ℝ}^n|{\forall }_i||v-x_i||>\epsilon \}.$$

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 $x^{\prime }\in {ℝ}^n$ 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 $A$. $\mathrm{\square }$