State the definition of the infimum of a set. For the set $$S = \left\{2(-1)^n+\frac{5}{n^2+2}: n \in \mathbb{N}^{+} \right\}$$

Find $\text{inf(S)}$ and prove your claim. Argue straight from the definition of infimum.

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 $k \le m$. But I'm unable to directly argue from it. Do you think what's presented below would get the marks, given this was from an exam?

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 $\min(-2, 2)<x_n < \max(11/3, -1/3)$ which implies that $-2<x_n < \dfrac{11}{3}$. Thus $\text{inf}(S) = -2.$ Does this count as arguing from definition?

If it doesn't I'd be grateful if someone could show me how to do this straight from the definition.


Clearly $-2$ is a lower bound for the set. We need to prove that $-2 + \epsilon$ is not a lower bound for the set for any $\epsilon > 0$. Indeed, if $\epsilon > 0$, then there is an odd integer $2n+1$ with $$\frac{5}{(2n+1)^2 + 2} < \epsilon.$$ But then $x_{2n+1} = \frac{5}{(2n+1)^2 + 2} - 2 < -2 + \epsilon.$ Hence $-2+\epsilon$ is not a lower bound for the set for any $\epsilon > 0$. But this implies that $-2$ is the greatest lower bound which by definition means that $-2$ is the infimum of the set.

Your analysis only shows that $-2$ is a lower bound, you didn't show that it is the greatest lower bound.

  • $\begingroup$ Thank you. Just to be clear, would my analysis count as showing that $-2$ is the greatest lower bound if the question didn't specify arguing from the definition, or is it that even then it would only amount to just showing that $-2$ is a lower bound? $\endgroup$ – user129566 Mar 21 '16 at 2:21
  • $\begingroup$ Well you never really mention that $-2$ is larger than any other lower bound so no, I don't think your analysis proves that $-2$ is the greatest lower bound, just a lower bound. $\endgroup$ – User8128 Mar 21 '16 at 2:44
  • $\begingroup$ Many thanks again. $\endgroup$ – user129566 Mar 21 '16 at 2:47

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