I'd like to show that operations on irrational numbers produce (most often) another irrational number. The approach I have in mind is to show that rational numbers make a cut $(A,B)$ and have $max(A)$ or $min(B)$, which is untrue for irrationals. And then from the bounds of these numbers reconsturct the resulting one. Is it worthwhile?
It has to be a limit point. But then again if it has to satisfy that"deleted $\delta$ neighborhood of $a$ contains points of $S$", then certainly "deleted neighborhood" implies that some point Is Not in $S$
@user681391 it is a limit point, for $\delta = 1$. However not the kind of point you asked for. Its neighbourhood will overlap with some other point in $(-1;1)$
Although, I would like to get back at my previous answer to your two examples. Perhaps there can be infinitely many neighbourhoods if we take arbitrary point of either set with arbitrary small $\delta$
A point $a \in S $ is called an accumulation point or limit point of $S$ if every deleted $\delta$ neighborhood of $a$ contains points of $S$. Can we regarard these accumulation points $a$ as complements to $S$?
Variables $x$ and $y$ don't appear before that in the paper. And they are introduced in these equations (marked by arrows) that are not foreshadowed themselves. I suspect they have to do with the narrowing of the bounds around $D$ but can't connect the two together. So, my question is: "Where the hell did that come from?"