COURSE TITLE – UNIT X
... straightedge, paper folding, tracing paper, mira, or computer to construct congruent segments, angles, triangles, and circles; an angle bisector; a perpendicular bisector; a perpendicular line from a point on a line; parallel lines; proportional segments; tangents; and inscribed and circumscribed po ...
... straightedge, paper folding, tracing paper, mira, or computer to construct congruent segments, angles, triangles, and circles; an angle bisector; a perpendicular bisector; a perpendicular line from a point on a line; parallel lines; proportional segments; tangents; and inscribed and circumscribed po ...
Knowledge space theory and union
... sets conjecture: “For any finite union-closed family of finite sets, other than the family consisting only of the empty set, there exists an element that belongs to at least half of the sets in the family.” (Frankl, 1979). The above is a conjecture made by Frankl in 1979; it is still an open problem ...
... sets conjecture: “For any finite union-closed family of finite sets, other than the family consisting only of the empty set, there exists an element that belongs to at least half of the sets in the family.” (Frankl, 1979). The above is a conjecture made by Frankl in 1979; it is still an open problem ...
Problem Sheet 2 Solutions
... 1. Let X be an infinite set and endow it with the cofinite topology τ . Is (X, τ ) Hausdorff? Is it compact? Solution: We will prove that X is not Hausdorff but it is compact. Suppose X is Hausdorff and let x, y ∈ X with x 6= y. Then there exist open sets U , V with x ∈ U , y ∈ V and U ∩ V = ∅. Sinc ...
... 1. Let X be an infinite set and endow it with the cofinite topology τ . Is (X, τ ) Hausdorff? Is it compact? Solution: We will prove that X is not Hausdorff but it is compact. Suppose X is Hausdorff and let x, y ∈ X with x 6= y. Then there exist open sets U , V with x ∈ U , y ∈ V and U ∩ V = ∅. Sinc ...
1. Compactness for metric spaces For a metric space (X, d) we will
... Corollary 1.6. Every compact metric space X is totally bounded, separable and second countable. Definition 1.7. Let (X, d) be a metric space. A Cauchy sequence xi ∈ X, i ∈ N is a sequence with the property that for every ε > 0 there is some i0 ∈ N such that d(xi , xj ) < ε for all i, j ≥ i0 . Every ...
... Corollary 1.6. Every compact metric space X is totally bounded, separable and second countable. Definition 1.7. Let (X, d) be a metric space. A Cauchy sequence xi ∈ X, i ∈ N is a sequence with the property that for every ε > 0 there is some i0 ∈ N such that d(xi , xj ) < ε for all i, j ≥ i0 . Every ...
MATH 161 SAMPLE FINAL EXAM SOLUTIONS 1. Euclidean geometry
... (Note: some books say “for some line ` and some point P not on `...,” so this version would also be acceptable, although it is not very obvious that the two versions are equivalent.) (2) A Saccheri quadrilateral is a quadrilateral ABCD such that AB = CD and the angles ∠B and ∠C are right angles. (No ...
... (Note: some books say “for some line ` and some point P not on `...,” so this version would also be acceptable, although it is not very obvious that the two versions are equivalent.) (2) A Saccheri quadrilateral is a quadrilateral ABCD such that AB = CD and the angles ∠B and ∠C are right angles. (No ...
Surveying Introduction
... Determining: both points already exist determine their relative locations. Establishing: one point, and the location of another point relative to the first, are known. Find the position and mark it. Most property surveys are re-surveys ...
... Determining: both points already exist determine their relative locations. Establishing: one point, and the location of another point relative to the first, are known. Find the position and mark it. Most property surveys are re-surveys ...
notes on the proof Tychonoff`s theorem
... C = {U ⇢ O : U covers X but does not admit a finite subcover }. If X is not compact, then C is non-empty. We will derive a contradiction in this case. First note that if C is partially ordered by inclusion ⇢ and if D is an ordered subset S of C, then U := D is an upper bound for D. To see this, we n ...
... C = {U ⇢ O : U covers X but does not admit a finite subcover }. If X is not compact, then C is non-empty. We will derive a contradiction in this case. First note that if C is partially ordered by inclusion ⇢ and if D is an ordered subset S of C, then U := D is an upper bound for D. To see this, we n ...