1. **State the problem:** We have the set $$A = \left\{ -1 + \frac{1}{2n + 1^2} : n \in \mathbb{N} \right\}$$ and we want to show that it is bounded. 2. **Show that $A$ is bounded:** - Since $n \in \mathbb{N}$, $n \geq 1$. - The term inside the set is $$-1 + \frac{1}{2n + 1^2} = -1 + \frac{1}{2n + 1}$$ - As $n$ increases, $2n + 1$ increases, so $\frac{1}{2n + 1}$ decreases. - The smallest value of $n$ is 1, so the largest element of $A$ is at $n=1$: $$-1 + \frac{1}{2(1) + 1} = -1 + \frac{1}{3} = -\frac{2}{3}$$ - As $n \to \infty$, $\frac{1}{2n + 1} \to 0$, so the elements approach $-1$ from above. - Therefore, all elements satisfy: $$-1 < -1 + \frac{1}{2n + 1} \leq -\frac{2}{3}$$ - Hence, $A$ is bounded below by $-1$ and bounded above by $-\frac{2}{3}$. 3. **Find $\sup(A), \inf(A), \max(A), \min(A)$:** - $\sup(A)$ is the least upper bound. Since the largest element is at $n=1$, $$\max(A) = \sup(A) = -\frac{2}{3}$$ - $\inf(A)$ is the greatest lower bound. The elements approach $-1$ but never reach it, so $$\inf(A) = -1$$ - Since $-1$ is not attained by any element (no $n$ makes the fraction zero), $$\min(A) \text{ does not exist}$$ 4. **Summary:** $$\boxed{\sup(A) = \max(A) = -\frac{2}{3}, \quad \inf(A) = -1, \quad \min(A) \text{ does not exist}}$$ This shows $A$ is bounded and identifies its bounds and extrema.