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MA2223: SOLUTIONS TO ASSIGNMENT 2 1. Determine which of the following subsets of R are open. (a) {x ∈ R : 1 < x < 3} Answer: Open. Reason: We proved in class that in a metric space every open ball is an open set. Here we have an open ball in R with centre 2 and radius 1. (b) S = {x ∈ R : |x| ≥ 1} Answer: Not open. Reason 1: The points −1 and 1 are contained in S but they are not interior points of S. Any open ball with centre 1 and radius r > 0 will contain the point 1 − where = min{1, 2r }. Since this point lies outside S, 1 is not an interior point. Similar argument for −1. Reason 2: The complement of S is the open interval (−1, 1) which is an open ball in R with centre 0 and radius 1 and hence is an open set. We proved in class that the complement of an open set is a closed set. This means S is a closed set. Since R is a connected space the only subsets of R which are both open and closed are ∅ and R. Hence S cannot be open. Reason 3: Let x be a limit point of S. Then there exists a sequence (xn ) in S which converges to x. Since the absolute value |.| : R → R is a continuous function, the sequence (|xn |) converges to |x|. Now |xn | ≥ 1 for all n and so from Real Analysis we know that |x| ≥ 1. Thus x ∈ S and so S is closed. As noted in Reason 2, this implies S is not open. (c) R\Z = {x ∈ R : x ∈ / Z} Answer: Open. 1 2 MA2223: SOLUTIONS TO ASSIGNMENT 2 Reason: We proved in class that the union of a collection of open sets is an open set. Here we have a union of open intervals (n, n + 1) where n ∈ N. (d) Q Answer: Not open. Reason: None of the points in Q are interior points. Let x ∈ Q be a rational number. Recall from Real Analysis that between any two real numbers there exists an irrational number. Thus every open interval (x−r, x+r) (i.e. every open ball in R with centre x) contains an irrational number. This shows x is not an interior point of Q. 2. Determine which of the following subsets of R2 are open. (a) {(x, y) ∈ R2 : (x − 3)2 < 4 − (y − 5)2 } Answer: Open. Reason: This is an open ball in R2 with centre (3, 5) and radius 2. Again we use the fact that in a metric space every open ball is an open set. (b) S = {(x, y) ∈ R2 : x > 0 or y > 0} Answer: Open. Reason 1: Write S = U ∪ V where U = {(x, y) ∈ R2 : x > 0} V = {(x, y) ∈ R2 : y > 0} If (a, b) ∈ U then the open ball B((a, b), a2 ) with centre (a, b) and radius a 2 is contained in U . Thus U is an open set. Similarly, if (a, b) ∈ V then the open ball B((a, b), 2b ) with centre (a, b) and radius b 2 is contained in V . Thus V is an open set. Now S is an open set (since it is a union of open sets). Reason 2: Write S = U ∪ V where U = {(x, y) ∈ R2 : x > 0} MA2223: SOLUTIONS TO ASSIGNMENT 2 3 V = {(x, y) ∈ R2 : y > 0} Let p : R2 → R, (x, y) 7→ x be the projection onto the first coordinate. Then p is a continuous map. Let W = {x ∈ R : x > 0} Note that W is an open set in R. Now U = p−1 (W ) and so U is an open set. Similar argument shows V is an open set. As in Reason 2 this implies S is open. (c) S = {(x, y) ∈ R2 : 1 < x < 3 and y = 0} Answer: Not open. Reason 1: None of the points in S are interior points. Let (x, 0) ∈ S. If B((x, 0), r) is an open ball in R2 with centre (x, 0) and radius r > 0 then B((x, 0), r) contains the point (x, 2r ). But (x, 2r ) is not contained in S and so (x, 0) is not an interior point of S. Reason 2: The complement of S is not closed. For example (2, 0) is a limit point of the complement of S since the sequence ((2, n1 ))∞ n=1 lies in the complement of S and converges to (2, 0). But (2, 0) ∈ S. 3. Determine which of the following subsets of R3 are open. (a) S = {(x, y, z) ∈ R3 : y 2 ≥ 1 − z 2 − x2 } Answer: Not open. Reason 1: Points which lie on the unit sphere S((0, 0, 0), 1) are contained in S but are not interior points of S. For example, consider the point (1, 0, 0) ∈ S. If B((1, 0, 0), r) is an open ball then the point (1 − , 0, 0) where = min{1, 2r } is contained in B((1, 0, 0), r) but not in S. This shows (1, 0, 0) is not an interior point of S. Reason 2: S is the complement of the open ball in R3 with centre (0, 0, 0) and radius 1. This means S is closed. Since R3 is connected, the only subsets which are both open and closed are ∅ and R3 . Hence S cannot be open. Reason 3: Let x ∈ R3 be a limit point of S. Then there exists a 4 MA2223: SOLUTIONS TO ASSIGNMENT 2 sequence (xn ) in S which converges to x. Since the Euclidean norm k.k : R3 → R is continuous, the sequence (kxn k) converges to kxk in R. Now kxn k ≥ 1 for all n and so from Real Analysis we know kxk ≥ 1. Hence x ∈ S and so S is closed. As in Reason 2 this implies S is not open. (b) S = {(x, y, z) ∈ R3 : x2 + y 2 < 1 and z = 0} Answer: Not open. Reason: None of the points in this set are interior points. Let (x, y, 0) ∈ S. If B((x, y, 0), r) is an open ball with centre (x, y, 0) then the point (x, y, 2r ) is contained in B((x, y, 0), r) but not in S. This shows (x, y, 0) is not an interior point of S. (c) S = {(x, y, z) ∈ R3 : x = cos t, y = sin t, z = t, t ∈ [0, 4π]} Answer: Not open. Reason 1: None of the points in this set are interior points. For example, consider the point (1, 0, 0) ∈ S. If B((1, 0, 0), r) is an open ball with centre (1, 0, 0) then the point (1, 0, ) where = min{π, 2r } is contained in B((1, 0, 0), r) but not in S. This shows (1, 0, 0) is not an interior point of S. Reason 2: S is the image of the continuous map T : [0, 4π] → R3 , t 7→ (cos t, sin t, t) Since [0, 4π] is compact (by Heine-Borel theorem), S is also compact. Hence S is closed (and bounded) in R3 . As R3 is a connected space, S cannot be both open and closed.