Exercise 588Topology › Topology of ℝⁿL1Added 3 Nov 2026

Limit points in $\mathbb{R}^n$

#limit points#sequences#closure

Statement

Show that $x$ is in the closure of $A\subseteq\mathbb{R}^n$ iff there is a sequence $(x_k)$ in $A$ with $x_k\to x$ .

Hint

Use $\varepsilon$ -balls and sequential characterisation.

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$\Leftarrow$ . If $x_k\to x$ with $x_k\in A$ , every ball around $x$ contains a $x_k$ . So $x$ is a limit point of $A$ ; $x\in\overline A$ .

$\Rightarrow$ . If $x\in\overline A$ , for each $k$ pick $x_k\in A\cap B(x,1/k)$ ; such a ball always meets $A$ by definition of closure.

Conclusion. Closure equals the set of limits of sequences in $A$ .