answer:To determine if S is simply connected, we need to check if it is path connected and if every closed curve in S can be contracted to a point. First, let's show that S is path connected. Given any two points (x1, y1) and (x2, y2) in S, we can find a path connecting them. If the line segment joining these two points does not pass through the origin, then this line segment itself serves as the required path. If the line segment does pass through the origin, then we can construct a path by going from (x1, y1) to (x1, 0) (or (0, y1) if x1 is 0) along a horizontal (or vertical) line, then moving along a semicircle centered at the origin with radius 1, and finally connecting to (x2, y2) along another horizontal (or vertical) line and then a line segment. Thus, S is path connected. However, S is not simply connected because there are closed curves that cannot be contracted to a point. Consider a circle centered at the origin with radius 1, denoted by C. Suppose there is a continuous deformation of C to a point within S, i.e., a homotopy H(t,θ) where t ∈ [0,1] and θ ∈ [0,2π] such that H(0,θ) = (cos(θ), sin(θ)) and H(1,θ) = (a,b) for some fixed point (a,b) in S, and H(t,θ) is continuous. Consider the function f(t) = ∫[0,2π] (x dy - y dx), where (x,y) = H(t,θ). By Stokes' theorem, f(t) is constant. For t = 0, this is the line integral of the vector field (x, -y) along the unit circle, which evaluates to 2π. For t = 1, since H(1,θ) is a constant, the line integral evaluates to 0, leading to a contradiction. Therefore, no such homotopy exists, and S is not simply connected. The fundamental group of S, denoted by π1(S), is isomorphic to the integers, reflecting the fact that closed curves in S can be classified by how many times they wrap around the removed origin.

answer:When I say that S is not simply connected because there are closed curves that cannot be contracted to a point, I mean that there are closed curves in S that cannot be continuously deformed to a point within S without tearing or popping the curve. Intuitively, imagine taking a rubber band and stretching it around the origin in S. As long as the rubber band doesn't pass through the origin, it represents a closed curve in S. Now, try to shrink the rubber band down to a point without moving it through the origin. You'll find that you can't do it without breaking the rubber band or moving it through the forbidden origin. Mathematically, this means that there is no continuous function, or homotopy, that transforms the closed curve into a point within S. In other words, there is no way to continuously deform the closed curve to a point without leaving S. In the case of S = R^2 - { (0,0) }, any closed curve that wraps around the origin cannot be contracted to a point because it gets stuck around the origin. This is in contrast to R^2, which is simply connected because any closed curve can be contracted to a point without obstruction. This property of S reflects the fact that removing a point from R^2 creates a hole that prevents some closed curves from being contracted to a point. This hole is not just a geometric feature, but a topological one, and it's what makes S not simply connected.

answer:To determine if A is simply connected, we need to check if it is path connected and if every closed curve in A can be contracted to a point within A. First, let's show that A is path connected. Given any two points (x1, y1) and (x2, y2) in A, we can find a path connecting them. We can construct a path by moving radially from (x1, y1) to the circle centered at the origin with radius √(x1^2 + y1^2), then move along this circle to the radial line passing through (x2, y2), and finally move radially to (x2, y2). This shows that A is path connected. However, A is not simply connected. Consider a circle centered at the origin with radius 3/2, denoted by C. This circle is contained in A, and it cannot be contracted to a point within A. If we try to shrink C, it will get stuck around the inner boundary of A (the circle with radius 1). More formally, we can use the same argument as before, using Stokes' theorem to show that there is no homotopy that contracts C to a point within A. Alternatively, we can use the fact that the fundamental group of A is isomorphic to the integers, which implies that A is not simply connected. The reason A is not simply connected is that it has a hole in the middle, which prevents some closed curves from being contracted to a point. This hole is not just the removed disk (i.e., the region { (x,y) in R^2 : √(x^2 + y^2) < 1 }), but also the fact that the inner boundary of A (the circle with radius 1) is not contractible within A.

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