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Find supremum and infimum for the sets:

$A=\{x^2+x+2: x \in \mathbb{R} \},$

$B=\{n^2+n-2: n \in \mathbb{Z} \}.$

I think that

$\sup_A=+\infty$ and $\inf_A=\frac{-7}{4},$

$\sup_B=+\infty$ and $\inf_B=-2.$

I found these values based on the graphs of these functions. It is correct? I should prove it somehow or my answer is enough?

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  • $\begingroup$ No its 1.75 and -2, the equation is a upward parabola, infimum and supremum of a parabola always exists on its vertex. $\endgroup$ Jan 12, 2020 at 13:17
  • $\begingroup$ Here its vertex is -0.5, substituting we get 1.75, since the parabola is upward it can not take the value below -2, so for integer solution we have -2. $\endgroup$ Jan 12, 2020 at 13:19

2 Answers 2

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$$x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\geq\frac{7}{4}.$$ The equality occurs for $x=-\frac{1}{2},$ which says that $$\inf_{x\in\mathbb R}(x^2+x+2)=\frac{7}{4}$$ and it's not $\frac{-7}{4}.$

The supremum of $A$ is indeed $+\infty$ because $$\lim_{x\rightarrow+\infty}(x^2+x+2)=\lim_{x\rightarrow+\infty}x^2\left(1+\frac{1}{x}+\frac{2}{x^2}\right)=+\infty.$$ Also, $$n^2+n-2=n(n+1)-2\geq-2,$$ which gives an infimum of $B$.

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If $f(x)=x^2+x+2$ then $f'(x)=2x+1$ which is $0$ when $x=-0.5$, for which $f(-0.5)=1.75$, so $\inf(A)=1.75$ and $f(x)$ is unbounded so supremum is $\infty$.

For $B$ let $g(x)=x^2+x-2$ then $g$ also has a minimum in $-0.5$. The closest integers to $-0.5$ is $-1$ and $0$ and the minimum for $g(n)$ when $n \in \Bbb Z$ must be found between these two values since it is a second order polynomial. One sees that $g$ obtain the same value for $-1$ and $0$ for which $g(0)=g(1)=-2$. So $\inf(B)=-2$ and $\sup(B)=\infty$ again

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