**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 .

In the discrete topological space (X,D), what is the closure of A where A is a subset of X? |

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.