Let $X$ be a normed space and $ B(0,R) := \{ x \in X : ||x|| < R \} \qquad \overline{B}(0,R) := \{ x \in X : ||x|| \le R \}. $ I want to show that $\overline{B(0,R)} = \overline{B}(0,R)$. I already proofed the inclusion $\overline{B(0,R)} \subseteq \overline{B}(0,R)$, for the other inclusion I consider $x \in \overline{B}(0,R)$, then $||x|| \le R$, if $||x|| = R$ then I have to show that every ball around $x$ contains some different point from $B(0,R)$, but how can I be sure? What if the normed space is discrete, then there are no limit points at all?
Show that the closure of an open ball is exactly $B(0,R) := \{ x \in X : ||x|| \le R\}$
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geometry
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functional-analysis
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0ahh okay i conjucted that something like that must be hold, but then i still does not know what ensures that every ball around $x$ contains another point? – 2012-11-09
1 Answers
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Assume that $X \neq 0$.
If $\lVert x \rVert = R$ and $\lambda \geq 0$ then $\lVert \lambda x\rVert = \lambda\lVert x \rVert = \lambda R$.
For $\lambda_n = 1-\frac{1}{n}$ we have $\lVert \lambda_n x \rVert \lt R$ and $\lambda_n x \to x$ because $\lVert x - \lambda_n x\rVert = \frac{1}{n}R \xrightarrow{n\to\infty} 0$. This shows that $X$ is not discrete because it contains a non-constant convergent sequence. Moreover, every ball around $x$ intersects $B(0,R)$ non-trivially: given $r \gt 0$ we can choose $n$ so large that $0 \lt \frac{R}{n} \lt r$ so that $\lambda_n x \in B(0,R) \cap B(x,r)$ because $\lVert \lambda_n x\rVert = (1-\frac{1}{n})R \lt R$ and $\lVert x - \lambda_n x\rVert = \frac{1}{n}R \lt r$.