Mid-Term Exam - Stony Brook Mathematics
... Suppose not. Then the open sets Uj = X − Aj cover X. Since X is compact, only finitely many are needed, so X = U1 ∪ · · · ∪ Un for some n. But since Uj ⊂ Uj+1 , this implies that X = Un . However, X − Un = An 6= ∅, so this is a contradiction. 5. Let U ⊂ R be an open set. Prove that U is a countable ...
... Suppose not. Then the open sets Uj = X − Aj cover X. Since X is compact, only finitely many are needed, so X = U1 ∪ · · · ∪ Un for some n. But since Uj ⊂ Uj+1 , this implies that X = Un . However, X − Un = An 6= ∅, so this is a contradiction. 5. Let U ⊂ R be an open set. Prove that U is a countable ...
Solutions to Midterm 2 Problem 1. Let X be Hausdorff and A ⊂ X
... }α∈Λ a family of compact subspaces, and U ⊂ X an open T X be a Hausdorff space, {Kα T set containing α∈Λ Kα . Prove that U contains α∈A Kα for some finite subset A ⊂ Λ. We can assume that X itself is compact (otherwise replace it with one of Kα ); hence, so is X r U . Since X is Hausdorff, all Kα ar ...
... }α∈Λ a family of compact subspaces, and U ⊂ X an open T X be a Hausdorff space, {Kα T set containing α∈Λ Kα . Prove that U contains α∈A Kα for some finite subset A ⊂ Λ. We can assume that X itself is compact (otherwise replace it with one of Kα ); hence, so is X r U . Since X is Hausdorff, all Kα ar ...
Manifolds
... be a homeomorphism. By composing fx with a translation of Rm , we may assume fx (x) = ~0. Let Ax = fx−1 (B(~0, 1)). The closure of Ax in X, Āx , is just the set fx−1 (closed ball B̄(~0, 1)), since that pre-image set is compact and hence closed in X. Similarly, let Bx = f −1 (B(~0, 2)). The set Bx i ...
... be a homeomorphism. By composing fx with a translation of Rm , we may assume fx (x) = ~0. Let Ax = fx−1 (B(~0, 1)). The closure of Ax in X, Āx , is just the set fx−1 (closed ball B̄(~0, 1)), since that pre-image set is compact and hence closed in X. Similarly, let Bx = f −1 (B(~0, 2)). The set Bx i ...
Topology, MM8002/SF2721, Spring 2017. Exercise set 3 Exercise 1
... • Show that if X and Y are first countable, so is X × Y . • Show that if X and Y are second countable, so is X × Y . Which of these statements hold for all finite products? Which for arbitrary products? Exercise 7. Consider an arbitrary product of topological spaces Πα∈A Xα . Show that the two bases ...
... • Show that if X and Y are first countable, so is X × Y . • Show that if X and Y are second countable, so is X × Y . Which of these statements hold for all finite products? Which for arbitrary products? Exercise 7. Consider an arbitrary product of topological spaces Πα∈A Xα . Show that the two bases ...
Document
... Prove that in a topological space ( X , ) the set A consisting of the elements of a convergent sequence along with its limit point is compact. ...
... Prove that in a topological space ( X , ) the set A consisting of the elements of a convergent sequence along with its limit point is compact. ...
The geometry of the universe - University of Maryland Astronomy
... the space is infinite, whereas for a spherical geometry the space is finite. As we will explore briefly in the next section this does not have to be the case, but it is in the simplest topology consistent with these geometries. What would this mean? First, let’s make something clear. In the standard ...
... the space is infinite, whereas for a spherical geometry the space is finite. As we will explore briefly in the next section this does not have to be the case, but it is in the simplest topology consistent with these geometries. What would this mean? First, let’s make something clear. In the standard ...
Geometry on Curved Surfaces Lab Exercise
... the same sets of points that you used for part 1. How do your lines compare to part 1? Can you draw more than one straight line between two points with this method? ...
... the same sets of points that you used for part 1. How do your lines compare to part 1? Can you draw more than one straight line between two points with this method? ...
Handout 1
... about the Klein bottle as two Möbius bands glued together along their boundary circles. (6) Let X = I 2 be the unit square with the equivalence relation, (t, 0) ∼ (1 − t, 1) and (0, t) ∼ (1, 1 − t) for all 0 6 t 6 1, gluing the opposite sides in pairs and reversing the orientation of both pairs. Th ...
... about the Klein bottle as two Möbius bands glued together along their boundary circles. (6) Let X = I 2 be the unit square with the equivalence relation, (t, 0) ∼ (1 − t, 1) and (0, t) ∼ (1, 1 − t) for all 0 6 t 6 1, gluing the opposite sides in pairs and reversing the orientation of both pairs. Th ...
