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Homework Statement .

Let ##X## be a nonempty set and let ##x_0 \in X##.

(a) ##\{U \in \mathcal P(X) : x_0 \in U\} \cup \{\emptyset\}## is a topology on ##X##.

(b) ##\{U \in \mathcal P(X) : x_0 \not \in U\} \cup \{X\}## is a topology on ##X##.

Describe the interior, the closure and the boundary of the subsets of ##X## with respect to each of these two topologies.

The attempt at a solution.

(a) Let ##A \subset X##. Suppose ##x_0 \in A##, then ##A## is open so ##A=A^{\circ}##. Now suppose ##x_0 \not \in A##. Then, for any ##S \subset A##, ##x \not \in S## so ##A## doesn't contain open sets appart from ##\emptyset##; from here one deduces ##A^{\circ}=\emptyset##.

If ##x_0 \not \in A##, then ##A## is closed, so ##A=\overline{A}##.

If ##x_0 \in A##, then ##A## is open and note that for any ##F## such that ##A \subset F##, ##x_0 \in F##, this means there isn't any closed set containing ##A## appart from ##X##, so ##\overline{A}=X##.

(b) If ##x_0 \not \in A##, then ##A## is open, so ##A=A^{\circ}##. If ##x_0 \in A##, then ##x \in A^{\circ}## iff there is ##U## open such that ##x \in U \subset A##. As ##x_0 \not \in U##, it is easy to see that ##A^{\circ}=A \setminus \{x_0\}##.

If ##x_0 \in A##, A is closed so ##\overline{A}=A##. If ##x_0 \not \in A##, then, the smallest closed set containing ##A## is ##A \cup \{x_0\}## so ##\overline{A}=A \cup \{x_0\}##.

I can describe the boundary of ##A## in (a) and (b) using the identity ##∂A=\overline{A} \setminus A^{\circ}## but I wanted to know if what I did up to now is correct.

Let ##X## be a nonempty set and let ##x_0 \in X##.

(a) ##\{U \in \mathcal P(X) : x_0 \in U\} \cup \{\emptyset\}## is a topology on ##X##.

(b) ##\{U \in \mathcal P(X) : x_0 \not \in U\} \cup \{X\}## is a topology on ##X##.

Describe the interior, the closure and the boundary of the subsets of ##X## with respect to each of these two topologies.

The attempt at a solution.

(a) Let ##A \subset X##. Suppose ##x_0 \in A##, then ##A## is open so ##A=A^{\circ}##. Now suppose ##x_0 \not \in A##. Then, for any ##S \subset A##, ##x \not \in S## so ##A## doesn't contain open sets appart from ##\emptyset##; from here one deduces ##A^{\circ}=\emptyset##.

If ##x_0 \not \in A##, then ##A## is closed, so ##A=\overline{A}##.

If ##x_0 \in A##, then ##A## is open and note that for any ##F## such that ##A \subset F##, ##x_0 \in F##, this means there isn't any closed set containing ##A## appart from ##X##, so ##\overline{A}=X##.

(b) If ##x_0 \not \in A##, then ##A## is open, so ##A=A^{\circ}##. If ##x_0 \in A##, then ##x \in A^{\circ}## iff there is ##U## open such that ##x \in U \subset A##. As ##x_0 \not \in U##, it is easy to see that ##A^{\circ}=A \setminus \{x_0\}##.

If ##x_0 \in A##, A is closed so ##\overline{A}=A##. If ##x_0 \not \in A##, then, the smallest closed set containing ##A## is ##A \cup \{x_0\}## so ##\overline{A}=A \cup \{x_0\}##.

I can describe the boundary of ##A## in (a) and (b) using the identity ##∂A=\overline{A} \setminus A^{\circ}## but I wanted to know if what I did up to now is correct.

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