Final Answer:
(a)In the subspace topology, a set U ⊂ Y is open in Y if and only if it is open in the larger space X that contains Y.
(b) In the subspace topology, a set U ⊂ Y is closed in Y if and only if it is closed in the larger space X that contains Y.
Step-by-step explanation:
(a) If U is open in X, then U is open in the subspace topology on (Y):
Let U be an open set in X. By definition of the subspace topology, an open set in Y is obtained as the intersection of Y with an open set in (X. Hence, U ∩ Y is open in Y, as it is the intersection of Y with an open set in X, making U ∩ Y = U open in Y as well.
If U is open in the subspace topology on Y, then U is open in X:
Assume U is open in Y under the subspace topology. By definition of the subspace topology, an open set in Y is of the form V ∩ Y, where V is an open set in X. Therefore, U = V ∩ Y for some open set V in X. This implies that U is an intersection of an open set in X and Y, making it open in X as well.
(b) If U is closed in X, then U is closed in the subspace topology on Y:
Let U be a closed set in X. For any closed set V in X, the intersection of (V) with (Y) V ∩ Y is closed in Y by the definition of the subspace topology. Hence, U ∩ Y = U being the intersection of a closed set in X with Y implies that U is closed in Y.
If U is closed in the subspace topology on Y, then U is closed in (X):
Assume U is closed in Y under the subspace topology. By definition, a closed set in Y is of the form (V ∩ Y), where V is closed in X. Therefore, (U = V ∩ Y) for some closed set V in X. This implies that U is an intersection of a closed set in X and Y, making it closed in X as well.
Therefore, both parts confirm that openness and closedness of a set in the subspace topology on Y are directly linked to the corresponding properties of that set in the larger space X containing Y.
Complete Question
Let X be a topological space, and let Y ⊂ X have the subspace topology.
(a) Let Y⊂X be an open subset of a topological space X. Show that a set U⊂Y is open in the subspace topology on Y if and only if U is open in X.
(b) Let Y⊂X be a closed subset of a topological space X. Show that a set U⊂Y is closed in the subspace topology on Y if and only if U is closed in X.