Use the field axioms a. 0 0,> 0; and if x < 0, 0,x...

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Use the field axioms a. 0 <1 b. to show that If x > 0,> 0; and if x < 0,¹ <0. c. If x ≥ 0, and if for every > 0,x ≤ , x = 0. 2. Let S be a nonempty subset of R and let k E R. Let kS = {ksls € S}. Prove a. If k> 0, then sup kS = k·sup S b. If k < 0, then sup kS = k· inf S 3. Find the interior and the boundary of each set. a. {:nEN} b. [1,4] U (4,5) c. {re Q:0<r<n} d. (SER\Q:s 2≥n} e. [0,3] n [3,4] 4. Show that compactness is necessary in the Nested Intervals Theorem. That is, find a family of intervals (An: n E N} such that An+1 An Vn and nn An = 0 and such that a. The sets A, are all closed. b. The sets A,, are all bounded. 5. Let X = R² and let d be the Euclidean metric. Define w: R² x R² → R by d(x, y) w(x, y) = (d(x,0) + d(y,0) if x, y and 0 are collinear otherwise This is sometimes called the Washington metric, because of the similarity to the streets of D.C. Let p = (3,4) and q = (-1,1). Draw the following neighborhoods: a. N(0; 1) b. N(p; 2) c. N(q; 2) Use the field axioms a. 0 <1 b. to show that If x > 0,> 0; and if x < 0,¹ <0. c. If x ≥ 0, and if for every > 0,x ≤ , x = 0. 2. Let S be a nonempty subset of R and let k E R. Let kS = {ksls € S}. Prove a. If k> 0, then sup kS = k·sup S b. If k < 0, then sup kS = k· inf S 3. Find the interior and the boundary of each set. a. {:nEN} b. [1,4] U (4,5) c. {re Q:0<r<n} d. (SER\Q:s 2≥n} e. [0,3] n [3,4] 4. Show that compactness is necessary in the Nested Intervals Theorem. That is, find a family of intervals (An: n E N} such that An+1 An Vn and nn An = 0 and such that a. The sets A, are all closed. b. The sets A,, are all bounded. 5. Let X = R² and let d be the Euclidean metric. Define w: R² x R² → R by d(x, y) w(x, y) = (d(x,0) + d(y,0) if x, y and 0 are collinear otherwise This is sometimes called the Washington metric, because of the similarity to the streets of D.C. Let p = (3,4) and q = (-1,1). Draw the following neighborhoods: a. N(0; 1) b. N(p; 2) c. N(q; 2)

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Here are the solutions to the problems using field axioms 1a 0 1 By the field axiom definition of 0 as an additive identity we have 0 1 1 By the trich... View the full answer