I'm trying to prove the following theorem, and I'm not sure my proof holds.
Theorem. Let $(X,d)$ be a metric space, $p\in X$, and $r >0$. Then $$E = \left\{x\in X\ |\ d(x,p)\leq r\right\}$$ is a closed subset of $X$.
Proof. $E$ is closed if and only if it contains all of its limit points. Let $\{x_n\} \subset E$ such that $x_n \to x$. We claim that $x \in E$. Since $x_n$ converges, for all $\epsilon > 0$, there exists an $N$, such that for all $n \geq N$, $d(x,x_n) < \epsilon.$
By the triangle inequality $d(p,x) \leq d(p, x_n) + d(x_n, x)$ and so $d(p,x) < r + \epsilon$ for all $\epsilon > 0$. So, at the least $d(p,x) \leq r$, and $x$ is contained in E. The proof is complete.
Is my bolded step logically wrong? Thanks!
$\endgroup$ 21 Answer
$\begingroup$[Disclaimer: This might be a little bit too nitpicky on my part.]
First $E$ is not a unit closed ball, it's an $r$-closed-ball if you like.
More important than that, your bolded step does not really add anything to the previous sentence, viz., we have $\forall a,b\in\mathbb{R}$:
$$a\leq b \iff \forall \varepsilon>0: a<b+\varepsilon.$$
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