MHB Does this count as a proof for the infimum?

[NoName3]

I want to know whether the following the counts as a proof that infimum of the set $S =
\left\{2(-1)^n+\frac{5}{n^2+2}: n \in \mathbb{N}^{+} \right\}$ is $\text{inf}(S) = -2$.

Let $A \subseteq X$, where $X$ is some ordered field. Then $\text{inf}(A)$ is $m \in X$ such that for any $x \in A$ we have $m \le x$ and for any number $k \in X$ such that $k \le x$ we have $k \le m$. I'm not sure as to whether what's below misses the second part of the definition. So, is it enough?

Let $\displaystyle x_n = 2(-1)^n+\frac{5}{n^2+2}$, then $\displaystyle x_{2n} = 2+\frac{5}{(2n)^2+2}$ and $\displaystyle x_{2n+1} = \frac{5}{(2n+1)^2+2}-2$.

Let $j = 2n$ then $\displaystyle x_{j} = 2+\frac{5}{j^2+2}> 2+\frac{5}{(j+1)^2+2} = x_{j+1}.$ Thus $x_{j} > x_{j+1}.$ Hence $x_{j}$ is strictly decreasing from $\dfrac{11}{3}$ to $2$. Similarly, let $k = 2n+1$ then $\displaystyle x_k = \frac{5}{k^2+2}-2 >\frac{5}{(k+1)^2+2}-2$$ = x_{k+1}.$ Hence $x_k$ is strictly decreasing from $-\dfrac{1}{3}$ to $-2$. Hence we have $\min(-2, 2)<x_n < \max(11/3, -1/3)$ which implies that $-2<x_n < \dfrac{11}{3}$. Thus $\text{inf}(S) = -2.$ Is this correct, do I've to use $\epsilon-\delta$ proof?

 

[Evgeny.Makarov]

Let $A \subseteq X$, where $X$ is some ordered field.

The concept of infimum is defined on any (partially) ordered set.

Then $\text{inf}(A)$ is $m \in X$ such that for any $x \in A$ we have $m \le x$ and for any number $k \in X$ such that $k \le x$ we have $k \le m$.

This is not stated very well. The scope of the universally quantified $x$ is the clause "we have $m \le x$". After that, the scope closes and $x$ is no longer defined. Therefore, the second occurrence of $x$, namely, in the phrase "such that $k \le x$", requires its own universal quantifier. Note also that the scope of this second quantifier does not last until the end of the sentence, but only includes "$k \le x$". Formally,
\[
m=\inf A\iff (\forall x\in X\; m\le x)\land (k\le x\implies k\le m)
\]
is incorrect because the occurrence of $x$ in $k\le x$ is undefined. Also,
\[
m=\inf A\iff (\forall x\in X\; m\le x)\land \forall x\in X\;(k\le x\implies k\le m)
\]
is incorrect because it has a different meaning. The correct statement is
\[
m=\inf A\iff (\forall x\in X\; m\le x)\land ((\forall x\in X\;k\le x)\implies k\le m).
\]
In words, it says, "$\inf A$ is $m \in X$ such that for any $x \in A$ we have $m \le x$ and for any number $k \in X$ such that $k \le x$ for all $x\in X$ we have $k \le m$".

The rest of the proof also has some flaws, so I'll write my variant. We have to prove $\forall x\in X\; (-2\le x)$ and $(\forall x\in X\;k\le x)\implies k\le -2$. For the first claim,
\[
-2\le 2(-1)^n<2(-1)^n+\frac{5}{n^2+2}.
\]
For the second one, suppose that $k\le x$ for all $x\in X$, but $k>-2$. Let $\varepsilon=k-(-2)>0$ and consider an odd $n>\sqrt{\dfrac{5}{\varepsilon}}$. Then
\[
n^2>\frac{5}{\varepsilon}\implies n^2+2>\frac{5}{\varepsilon}\implies \frac{5}{n^2+2}<\varepsilon.
\]
Therefore,
\[
X\ni 2(-1)^n+\dfrac{5}{n^2+2}=-2+\dfrac{5}{n^2+2}<-2+\varepsilon=k.
\]
We found an $x\in X$ such that $x<k$, which contradicts the assumption that $x\ge k$ for all $x\in X$.

 

[NoName3]

Thank you, very helpful!