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I have to find $\sup$, $\inf$ and $\partial A$ of the set

$$ A = \left\{ (-1)^n \frac{n^2+1}{2n^2+3}:\ n\in\mathbb{N} \right\}$$

So as $n\to +\infty$ we oscillate between $\pm \frac{1}{2}$. Those are its limit points, but they do not belong to $A$, hence $A$ cannot be closed. This means that whatever is its boundary is not (entirely) contained in $A$.

I have managed to prove that $$\bigg| \frac{n^2+1}{2n^2+4}\bigg| \leq \frac{1}{2}$$

hence the set is bounded, because it does exist an upper bound and a lower bound.

I am not sure how to prove that $\frac{1}{2}$ is also the supremum (but it's not the max since it does not belong to $A$).

So I thought to define $a_n = (-1)^n \frac{n^2+1}{2n^2+3}$ and hence $\forall \epsilon > 0$, $\exists n_0$ such that $a_n \geq \frac{1}{2}-\epsilon$.

Say $n_0$ is even, hence $n_0 = 2k_0$ for some $k_0$. Then

$$a_{2k_0} = \frac{4k_0^2+1}{8k_0^2 + 3} \geq \frac{1}{2}-\epsilon$$

This means $k_0^2 > \frac{1}{16\epsilon} - \frac{3}{8}$.

Here I got stuck because I'm still incompetent and I cannot interpret and give sense to this result, assuming it has any...

  • On the other side, I have totally no clue about how to find the boundary...

Any help? Thank you!

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  • $\begingroup$ Use long division to rewrite the fraction as $(-1)^n\left (\frac 12 - \frac {1}{2(2n^2+3)} \right )$. I believe you'll find it significantly easier to analyze the expression's asymptotic behavior in this form. $\endgroup$ Nov 20 at 21:11
  • $\begingroup$ Another approach for the same idea: Try consider subsequences on odds and evens. One will be Rise to -1/2 second lower to 1/2. $\endgroup$ Nov 20 at 21:13
  • $\begingroup$ @RobertShore I already studied the behavious at infinity and I found its limit points... Are they its boundary too? How can I prove this? It's not always true that limit points are bondary points, is it? $\endgroup$
    – Numb3rs
    Nov 20 at 21:14
  • $\begingroup$ In this form, you can readily see that $\vert a_n \vert \lt \frac 12$ and that $\vert a_n \vert$ is monotonically increasing. That should help you determine the boundary. If $\vert x \vert \gt \frac 12$, can you necessarily find an open ball that's disjoint from the sequence? $\endgroup$ Nov 20 at 21:20
  • $\begingroup$ @RobertShore Can I also say the set is not closed because not every convergent sequence converge to elements in the set, like $x_n = \frac{1}{2} + n$? Anywy, I think it's boundary is $\{ -1/2, 1/2\}$ $\endgroup$
    – Numb3rs
    Nov 20 at 21:29

2 Answers 2

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Observe that $a\leq\frac12$ for every $a\in A$. This tells us that $\frac12$ is an upper bound of $A$.

Secondly observe that for every $r<\frac12$ we can show that $r$ is not an upper bound of $A$. This because an element $a\in A$ can be found with $r<a$.

This together proves that $\frac12$ is the least upperbound (i.e. supremum) of $A$.

In a similar way it can be proved that $-\frac12$ is the greatest lowerbound (i.e. infinum) of $A$.

Observe that the closure of $A$ is the set $A\cup\left\{-\frac12,\frac12\right\}$ and observe that the closure of $A^c=\mathbb R-A$ is $\mathbb R$.

Then for the boundary of $A$ we find:$$\partial A=\overline A\cap\overline{A^c}=A\cup\left\{-\frac12,\frac12\right\}$$

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  • $\begingroup$ That was really enlightening!! Thank you so much! $\endgroup$
    – Numb3rs
    Nov 21 at 10:32
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    $\begingroup$ Let me say one thing more about isolated points of a set $A$. In another question you wondered whether they are elements of the boundary of $A$. Mostly they do but not always. If we deal with discrete topology then the boundary of every set is empty. But next to that every point of a set can be recognized as an isolated point of that set. $\endgroup$
    – drhab
    Nov 21 at 10:46
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For even $n=2k$, $k\in\mathbb{N}$, note $$ (-1)^n \frac{n^2+1}{2n^2+3}=\frac{4k^2+1}{8k^2+3}=\frac12\frac{k^2+\frac14}{k^2+\frac38}<\frac12 $$ and $$ \lim_{k\to\infty}\frac{4k^2+1}{8k^2+3}=\frac12. $$ Also $$ \bigg(\frac{4x^2+1}{8x^2+3}\bigg)'=\frac{8x}{(8x^2+3)^2}>0, x\ge1. $$ Thus $\bigg\{\frac{4k^2+1}{8k^2+3}\bigg\}$ tends strictly increasingly to $\frac12$ and hence for $\forall \varepsilon>0$, $\exists k_0\in\mathbb{N}$ such that $$ \frac12-\varepsilon<\frac{4k_0^2+1}{8k_0^2+3}<\frac12 $$ or $$ \frac12-\varepsilon<a_{2k_0}<\frac12. $$ So $\sup A=\frac12$. Note $\frac12$ is a boundary point since there are points in $A$ and not in $A$ in any neighborhood of $\frac12$. You can use the same way to show $\inf A=-\frac12$ and $-\frac12$ is a boundary point of $A$ too.

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  • $\begingroup$ Can I say the boundary of $A$ is $A$ itself except for the points $\pm 1/2$, hence it's not closed? $\endgroup$
    – Numb3rs
    Nov 20 at 21:48
  • $\begingroup$ The boundary of $A$ is $\{\pm\frac12\}$. All points of $A$ are dangling points. $\endgroup$
    – xpaul
    Nov 20 at 22:08
  • $\begingroup$ What is a "dangling point"? o.O $\endgroup$
    – Numb3rs
    Nov 20 at 22:14
  • $\begingroup$ See the update. $\endgroup$
    – xpaul
    Nov 20 at 22:15
  • $\begingroup$ Thank you. Though I dont' understand why other points in $A$ are not boundary points too. For example for $n = 0$ (we assume it in the naturals) we have $1/3$. Any neighbourhood of $1/3$ contains points in $A$ and not in $A$, for example $\frac{1}{3} + \frac{1}{\sqrt{127}}$ $\endgroup$
    – Numb3rs
    Nov 20 at 22:17

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