Question

1.С., U element of A; this indicates, ACUU By double containment. A U Ur. Since A = U Ur this means A is the union of open sets. Which indicates A is open. Problem 13.4 part(a). (a) If {名} is a family of topologies on X, show that topology on X. Is Ua a topology on X? Solution: a) Show that n.名is a topology on X b.)Is Ua a topology on X? ,名is a a. Show that na is a topology on X Proof a: Since we claim , for α E A is a topology, by definition, of a topology we know , X are in 3α for every α Ε A, we have ø.x e oeA"7" [condition 1 satisfied If {Ug)аев NEA Since we clain . , for α E A is a topology if Use BUg E T, for α E A then U0EBUR E nEA3a. [condition 2] For the last condition, we know that if Un , Un E N0EA"7, we have U1, . . . , Un e Tx for α E A. But since these are topologies b. Is UT, a topology on X?

is 13.4 part (a) correct.

Answer 1

"(a) Your proof of ""Intersection F_2 is a topology"" is correct only. (b) In general, Union F_2 is not a topology. example: Let X = {a, b, c}. F_1 = {phi, X, {a, c} and F_2 = {phi, X, {a, b}. Clearly F_1 & F_2 are topologies on X. Now, F_1 Union F_2 = {phi, X, {a, c}, {a, b} is not topology because {a, c} Intersection {a, b} = {a} Notelementof F_1 Union F_2. I hope that it is clear. If still you have any doubt, please do comment. I would be very happy to reply you back."