Let X be an uncountable set and let tau = {Phi} {U X: U^c is countable}. Show th
ID: 3108488 • Letter: L
Question
Let X be an uncountable set and let tau = {Phi} {U X: U^c is countable}. Show that tau is a topology on X. Is it Hausdorff? Let A be a proper subset of X and closed. Show that all subsets of A closed. Prove that the countable intersection of open sets in T remains open result true in usual R.? Show that the finite intersection of open sets is non-empty. Is this ref usual R? Is Q= R? for the topology T. What about Q^c or [0, 1] ? Give an example of a set X so that T reduces to the discrete topologExplanation / Answer
1)the empty set belongs to tau.
2)the complement of X is the empty set and empty set is countable. therefore, X belongs to tau.
3)let{V_a|a in I} be any subfamily of tau. X- union of {V_a|a in I} = intersection of {X-V_a|a in I}. Since V_a is in tau X-V_a is countable for each a. Therefore, intersection of {X-V_a|a in I} is countable. that is, X- union of {V_a|a in I} is countable. that is, union of {V_a|a in I} belongs to tau.
4) let V1, V2,,,,,Vk be any finitely many elements in tau. X-Vi is countable for each i. Therefore, union of X-Vi is countable, that is, X -intersection of Xi is countable, i.e., intersection Vi belongs to tau.
Hence tau is a topology.
2) it is not hausdroff. let x and y be any two distict points in X. If U and V are disjoint open sets containing x and y respectively, since V is contained in X-U, V is countable and X-V is uncountable.
3) let A be a proper subset of X and let A be closed. Then X-A is open. then X-(X-A)=A is countable.Let B be any subset of A. B is countable. X-B is open. hence B is closed.
4)intersection of the family{(1-(1/n), 2+1/n)|n is a positive integer} is the open interval (1,2)
5) let V1, V2,,,,Vk be finitely many open sets. X-Viis countable for each i. union X-Vi is countable,i.e., X-intersection Vi is countable If intersection Vi is empty then X becomes countable which contradicts the given thing that X is uncountable
6) the set of irrational numbers is an open set in tau. so the set of rational numbers is closed set in tau and hence its closure is the set of rational numbers itself and not the set of real numbers. But the set of irrational numbers is dense. [0,1] is also dense.
7) let X be a countable set
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