# Find supremum and infimum for the sets: $A=\{x^2+x+2: x \in \mathbb{R} \},$ $B=\{n^2+n-2: n \in \mathbb{Z} \}.$

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?

• No its 1.75 and -2, the equation is a upward parabola, infimum and supremum of a parabola always exists on its vertex. Jan 12, 2020 at 13:17
• 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. Jan 12, 2020 at 13:19

$$x^2+x+2=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\geq\frac{7}{4}.$$ 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 $$B$$.
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