Final answer:
Ty = {Y ∪ U: U ∈ T} satisfies the topology axioms and is therefore a topology on Y. This is proven by confirming that Ty contains both Y and the empty set, is closed under arbitrary unions, and is closed under finite intersections.
Step-by-step explanation:
To show that the collection Ty = {Y ∪ U: U ∈ T} is a topology on Y, we need to verify that Ty satisfies the topology axioms. They are: (1) Y and the empty set are in Ty, (2) any union of elements of Ty is also in Ty, and (3) any finite intersection of elements of Ty is also in Ty.
Firstly, since Y is a subset of X and U can be X itself in T, Y ∈ Ty. Similarly, the empty set is included in any topology, hence the empty set is also in Ty. Secondly, for any collection {U_i} where U_i ∈ T, the union ∪ (Y ∪ U_i) = Y ∪ (∪ U_i), and since the union of any collection of open sets in T is also open in T, this union is in Ty. Lastly, for any finite collection {U_i}, the intersection ∩ (Y ∪ U_i) = Y ∪ (∩ U_i), and since the intersection of any finite number of open sets in T is also open in T, this intersection is in Ty.
Therefore, Ty fulfills all the criteria to be a topology on the subset Y of X, which confirms the assertion that Ty is indeed a topology. This concludes the proof that Ty is a topology on Y.