# Sup and inf and boundary of this set

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!

• 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. Nov 20 at 21:11
• 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. Nov 20 at 21:13
• @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? Nov 20 at 21:14
• 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? Nov 20 at 21:20
• @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 $\{ -1/2, 1/2\}$ Nov 20 at 21:29

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.

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:$$\partial A=\overline A\cap\overline{A^c}=A\cup\left\{-\frac12,\frac12\right\}$$

• That was really enlightening!! Thank you so much! Nov 21 at 10:32
• 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. Nov 21 at 10:46

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 $$\frac12-\varepsilon So $$\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.

• Can I say the boundary of $A$ is $A$ itself except for the points $\pm 1/2$, hence it's not closed? Nov 20 at 21:48
• The boundary of $A$ is $\{\pm\frac12\}$. All points of $A$ are dangling points. Nov 20 at 22:08
• What is a "dangling point"? o.O Nov 20 at 22:14
• See the update. Nov 20 at 22:15
• 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 $\frac{1}{3} + \frac{1}{\sqrt{127}}$ Nov 20 at 22:17