Topology on a set ##X## (find interior, closure and boundary of sets)

[mahler1]

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.

 

[HallsofIvy]

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 $\overline{A}=X$.

Are you thinking that "If U is open then it cannot be closed"? That is NOT in general true. There exist topologies in which there exist sets that are both open and closed and, indeed, a topology in which all sets are both open and closed.

(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.

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Fredrik]

I don't see anything wrong, but I agree with Halls that your explanation of why $\overline A=X$ in part (a) sounds a little bit like "since it it's open, it can't be closed". So you may want to clarify that if F≠X, then F isn't closed because its complement isn't open.

 

[mahler1]

I don't see anything wrong, but I agree with Halls that your explanation of why $\overline A=X$ in part (a) sounds a little bit like "since it it's open, it can't be closed". So you may want to clarify that if F≠X, then F isn't closed because its complement isn't open.

I know that there exist clopen sets so I don't know what I was thinking when I wrote that. I would like to know if now it is well justified: suppose $F \neq X$, since $x_0 \not \in F^c$, then $F^c$ is not open, which implies ${F^c}^c=F$ can't be closed.

 

[Fredrik]

Yes, that works.