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Let $F \subset \mathbb{R}^2$. Show that if $\forall R >0$, $B(0,R)\cap F$(open ball) is finite, then $F$ is discrete. Then show that if $F$ is closed, then the converse is true.

My attempt show the first part:

Let $F \subset \mathbb{R}^2$. Suppose $\forall R >0, B(0,R)\cap F$ is finite. Let $x \in F$. Then we have $B(x, R-||x||_2)\cap F \subset B(0, R) \cap F$. Thus $B(x, R-||x||_2)\cap F$ is finite. Suppose that the cardinal of $B(x, R-||x||_2)\cap F$ is $n \in \mathbb{N}$. We can enumerate its elements $B(x, R-||x||_2)\cap F = \{x, e_2, e_3, \dots, e_n\}$. Let's then put $\delta = min \{ ||x-e_i||_2 | i \in \{2, \dots,n\} \}$. Then $B(x, \frac{\delta}{2})\cap F = \{x \}$. Thus $F$ is discrete.

Is my proof correct? And I don't know how to reason in case $F$ is closed and show that the converse is true.

John Mayne
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1 Answers1

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If $B(0, R)\cap F$ were not finite for some $R>0$, then this is a bounded infinite subset of $\mathbb{R}^2$ and so has a limit point by Bolzano-Weierstrass Theorem. But $F$ is closed and so this limit point belongs to $F$. This contradicts that $F$ is discrete.

Prajwal Kansakar
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