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

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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.
 
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  • #2
HallsofIvy
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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##.
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





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The Attempt at a Solution

 
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  • #3
Fredrik
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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.
 
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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.
 
  • #5
Fredrik
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Yes, that works.
 

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