Talk:Zero is Limit Point of Integer Reciprocal Space

In the second case of the proof, it is assumed that $d$ must exist, which is not the case for negative numbers. --kc_kennylau (talk) 20:34, 28 October 2016 (EDT)

Trivial to fix, just add the case where $x < 0$ as a fourth case. --prime mover (talk) 03:09, 29 October 2016 (EDT)

Also, in the third case of the proof, it is (mistakenly) assumed that $b$ must be positive. --kc_kennylau (talk) 20:47, 28 October 2016 (EDT)

$U$ contains $0$ so $b > 0$. --prime mover (talk) 03:09, 29 October 2016 (EDT)

$U$ contains $0$ but $I$ does not necessarily contain $0$. It was not specified. --kc_kennylau (talk) 03:22, 29 October 2016 (EDT)

Again simple to fix, either construct $I$ so that $0 \in I$, or take the two cases where $0 \in I$ (in which case the proof continues as it is), and $0 \notin I$ (in which case it's not a set with an element of $A$ in it).