In discrete topological Space (X,D), the closure of A where A⊆X is the set A itself because the closure of the subset of X is the smallest closed super set of subset and in discrete topological Space (X,D), the smallest closed super set of A is the subset A itself .
Related Questions;
What is the closure of a subset of a topological space?
Since in case of discrete topological Space every subset of X is present in (X,D), So every subset of X is both open and closed that's why A is also both open and closed. As the Closure of subset of X is the intersection of all closed supper sets of Subset or is the smallest closed super set of subset and in discrete topological Space (X,D), the only set A is the smallest closed super set of itself, so in discrete topological Space (X,D) the closure of A where A is the subset of X is the set A itself.
Examples:
1) Let X = {9, 10, 11} and A = {10,11}
Cl(A) = ?
Solution;
D = { P(X) } where D denote discrete topology on X.
Since
P(X) = { Φ, X, {9}, {10}, {11}, {9,10}, {9,11}, {10,11} }
Now all closed sets are
Φ, X, {9}, {10}, {11}, {9,10}, {9,11}, {10,11}
All Closed super Sets of A are
X, {10,11}
Cl(A) = X ∩ {10,11} = {10,11}
Hence proved that Cl(A) is the Set A itself.
2) Let X = {d, e, f } and B = {d, f}
Cl(B) = ?
Solution;
D = { P(X) } where D is discrete topology on X.
Since
P(X) = { Φ, X, {d}, {e}, {f}, {d, e}, {d, f}, {e, f} }
All Closed Sets are
Φ, X, {d}, {e}, {f}, {d, e}, {d, f}, {e, f}
All Closed super Sets of B are
X, {d, f}
Cl(B) = X ∩ {d, e}
Hence proved that Cl(B) is the set B itself.