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Hint Needed for a Decreasing Sequence of Limit Points

Mathematics Asked on February 19, 2021

I’m trying to work through exercise 13E, part (4) of S. Willard’s Topology through self-study. I’m scratching my head on the question below. I paraphrased of some notation earlier in the exericse, which is required to understand the question.

Given a set $A$, let $A’$ denote the set of accumulation points of $A$. Let $A^1 = A’, A^2 = (A^1)’, A^3 = (A^2)’$ and so on . . . .

For any positive integer $n$, there is a set $A subseteq mathbb{R}$ such that $A,A^1,ldots,A^{n-1}$ are non-empty and $A^n = emptyset$.

Of course, we assume $mathbb{R}$ has the standard topology.

I’m scratching my head trying to construct such a set $A$. I’ve thought about taking the union of some collection of sets, each in a form similar to ${ 1/n : n in mathbb{N} }$, and arranging the limit points thereof so they fall off one by one as I continue to find the limit points of the limit points. I haven’t found a way to make this work, though, and I have a feeling I’m heading in the wrong direction.

Can someone please provide me a hint or some guidance? I’d like to work this out myself, so please do no provide a solution (or if you do, please conceal it as a spoiler). I appreciate your assistance.

Best,
Doug

One Answer

Here’s a schematic of what you need to get the case $n=4$ in the plane; it generalizes easily and also adapts fairly easily to $Bbb R$.

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Answered by Brian M. Scott on February 19, 2021

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