Find the accumulation points $(A^\prime)$ of $A = \mathbb{N}=\{ 1,2,3,4,\dots \} \subset \mathbb{R}$.

My Attempt:

Consider the following sketch of $A$:

enter image description here

  • $\underline{x < 1}:$

Consider $r^* =\frac{1}{2}|1-x|$. $B^\prime (x,r^*)\cap A = \emptyset \implies x \not\in A^\prime$.

  • $\underline{x = n, n\in \mathbb{N}}:$

Consider $r = \frac{1}{p}, p>1$. Clearly $0< r < 1$ and hence $B^\prime (x,r) \cap A = \emptyset \implies n \not\in A^\prime$

  • $\underline {n < x < n+1}:$

Let $R^* = \frac{1}{2}\min{\{|n-x|,|(n+1)-x| \}}$. Clearly $B^\prime (x,r^*) \cap A = \emptyset \implies x \not \in A^\prime$

We have thus shown that, if $A= \mathbb{N} \subset \mathbb{R}$, then $$A^\prime = \emptyset$$

Is this correct? Or am I missing anything?

  • $\begingroup$ It's correct. My only concern is that it's not made explicit how to choose $p$ in the second case, and not completely clear why that's necessary. $\endgroup$ – Jonathan Y. Apr 7 '15 at 20:17
  • $\begingroup$ @JonathanY. The distance between any two natural numbers that follow immediately after one another, is $1$ unit. So in order to find a radius so that the reduced neighborhood's interception with $A$ is the emptyset, we must have that the radius of the reduced neighborhood should be less than $1$. I defined this is $r = \frac{1}{p}, p > 1$ as this will ensure $r$ is less than $1$. How can I write that in a more mathematical way to give as "reason" in the example? $\endgroup$ – DJS Apr 7 '15 at 20:23
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    $\begingroup$ Yes. How do you choose $p$, though? If by the fact that $(1,\infty)\neq\emptyset$, why not choose $r$ by the same property of $(0,1)$? Or if you simply choose $p=2$, for example, why not take $r=0.5$? My point is, it's not clear what purpose $p$ serves. $\endgroup$ – Jonathan Y. Apr 7 '15 at 20:27
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    $\begingroup$ @JonathanY. Thank you! :). That makes sense. It's "redundant" having them choose $r = \frac{1}{p}$, since choosing only one $r$, such as $r= 0.5$ will suffice. We need only show there is one $r$, and hence that should be enough (and much less work for a reader). :). Thank you :) $\endgroup$ – DJS Apr 7 '15 at 20:31

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