(a) Show that A is closed:A set is said to be closed if it contains all of its limit points. It is equivalent to saying that a set is closed if it contains all of its accumulation points. A set that is not closed is called open.The set A = x² + y² ≤ 4,0 ≤ z ≤ 4 − x² - y² can be expressed as the intersection of the paraboloid z = 4 − x² - y² and the cylinder x² + y² = 4. Now we can prove A is closed, which is equivalent to demonstrating that it includes all of its limit points.Let us assume that (x₀, y₀, z₀) is a limit point of A. If we can show that (x₀, y₀, z₀) is a point of A, we will have shown that A contains all of its limit points.Among other things, the following are implied by the condition (x, y, z) ∈ A:-4 ≤ x ≤ 4,-4 ≤ y ≤ 4,0 ≤ z ≤ 4 − x² - y².(x, y, z) is a limit point of A if and only if it satisfies all of the following conditions:-4 ≤ x ≤ 4,-4 ≤ y ≤ 4,0 ≤ z ≤ 4 − x² - y².And for each ε > 0, there is a point (x, y, z) ≠ (x₀, y₀, z₀) such that the Euclidean distance between (x, y, z) and (x₀, y₀, z₀) is less than ε (i.e., the point (x, y, z) is in the ε-neighborhood of (x₀, y₀, z₀). The distance between (x, y, z) and (x₀, y₀, z₀) is defined as follows:d = sqrt((x - x₀)² + (y - y₀)² + (z - z₀)²).Because ε > 0, d > 0. Then there exists a point (x, y, z) in A for which d < ε. There are two cases to consider in order to finish the proof:Case 1: If 0 < z₀ ≤ 4, then there exists a positive ε such that the 3D ball Bε((x₀, y₀, z₀)) of radius ε around (x₀, y₀, z₀) lies inside A. It is because there is a positive number δ for which Bδ((x₀, y₀, z₀)) lies in the intersection of z = 4 − x² - y² and cylinder x² + y² ≤ 4, and we can choose ε as the smaller of δ and z₀.Case 2: If z₀ = 0, then we must choose ε < 1. The reason for this is because there is an infinite sequence of points in A that converge to (x₀, y₀, z₀). The sequence is defined as follows:(x₁, y₁, z₁) = ((1/2)ε, (1/2)ε, 0),(x₂, y₂, z₂) = ((2/3)ε, (2/3)ε, ε/3),(x₃, y₃, z₃) = ((3/4)ε, (3/4)ε, ε/2),...,(xₙ, yₙ, zₙ) = ((n/(n + 1))ε, (n/(n + 1))ε, ε/(n + 1)),...Then, the point (x₀, y₀, z₀) is an accumulation point of the sequence. As a result, we have shown that A contains all of its limit points, implying that A is closed.(b) Show that A is bounded:A set is said to be bounded if it is contained in some ball of finite radius. In other words, a set A is bounded if there exists a positive real number r such that A is contained in the ball of radius r centered at the origin. A set that is not bounded is said to be unbounded.A is contained within the cylinder x² + y² ≤ 4, as well as above the plane z = 0 and below the plane z = 4 - x² - y², among other things. The upper surface of A is clearly bounded, since it lies within a circle of radius 2 and is parallel to the xy-plane. As a result, we must show that the bottom surface is bounded as well.Let (x, y, 0) be a point on the xy-plane. We will demonstrate that the point lies within a disk of radius 2, centered at the origin and lying on the xy-plane.The formula x² + y² ≤ 4, which describes the cylinder x² + y² ≤ 4, guarantees that (x, y) lies inside a circle of radius 2 centered at the origin. Furthermore, the formula 0 ≤ z ≤ 4 - x² - y² implies that z ≤ 4. As a result, (x, y, 0) is contained in a disk of radius 2 centered at the origin and lying on the xy-plane, with height bounded above by 4. As a result, the set A is bounded.(c) Show that there is a point (x₀, y₀, z₀) in A such that f(x,y,z) ≤ f(x₀,y₀,z₀) for all (x,y,z) in A:The function f(x, y, z) = x + 3y² + z is a continuous function defined on a closed, bounded set A. As a result, by the Extreme Value Theorem, there must be a point (x₀, y₀, z₀) in A at which f(x, y, z) is minimal.Therefore, for all (x, y, z) in A, we have:f(x, y, z) ≤ f(x₀, y₀, z₀).