Mixing
The complement of an open set?
$begingroup$
The definition of closed set says that the complement of an open set is a closed set.
I found one theorem and its proof, and one step confused me.
We have to prove that one $x_1$ is unique on some interval $[t_0,t]$. So, we take another $x_2$. Then we have two separate sets:
Set $A=left { s in [t_0,t] mid x_1 (s) neq x_2(s) right }$
Set $A^C=left { s in [t_0,t] mid x_1 (s) = x_2(s) right }$
We want to prove that $A^C=[t_0,t]$.
One thing that confuses me is that they proved that both of $A$ and $A^C$ are open sets. I won't write that proof here, but, they all make perfect sense. How can two complements be both open?
general-topology
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
asked Jan 20 at 12:34
user25568user25568
1138
$endgroup$
add a comment |
$begingroup$
The definition of closed set says that the complement of an open set is a closed set.
I found one theorem and its proof, and one step confused me.
We have to prove that one $x_1$ is unique on some interval $[t_0,t]$. So, we take another $x_2$. Then we have two separate sets:
Set $A=left { s in [t_0,t] mid x_1 (s) neq x_2(s) right }$
Set $A^C=left { s in [t_0,t] mid x_1 (s) = x_2(s) right }$
We want to prove that $A^C=[t_0,t]$.
One thing that confuses me is that they proved that both of $A$ and $A^C$ are open sets. I won't write that proof here, but, they all make perfect sense. How can two complements be both open?
general-topology
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
asked Jan 20 at 12:34
user25568user25568
1138
$endgroup$
2
$begingroup$
A set can be both closed and open or neither, they're not each other's opposite
$endgroup$
– Alessandro Codenotti
Jan 20 at 12:36
1
$begingroup$
When one is empty, for instance. Or you can consider “disconnected” spaces, such as ${0,1}$.
$endgroup$
– Mindlack
Jan 20 at 12:37
add a comment |
$begingroup$
The definition of closed set says that the complement of an open set is a closed set.
I found one theorem and its proof, and one step confused me.
We have to prove that one $x_1$ is unique on some interval $[t_0,t]$. So, we take another $x_2$. Then we have two separate sets:
Set $A=left { s in [t_0,t] mid x_1 (s) neq x_2(s) right }$
Set $A^C=left { s in [t_0,t] mid x_1 (s) = x_2(s) right }$
We want to prove that $A^C=[t_0,t]$.
One thing that confuses me is that they proved that both of $A$ and $A^C$ are open sets. I won't write that proof here, but, they all make perfect sense. How can two complements be both open?
general-topology
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
asked Jan 20 at 12:34
user25568user25568
1138
$endgroup$
The definition of closed set says that the complement of an open set is a closed set.
I found one theorem and its proof, and one step confused me.
We have to prove that one $x_1$ is unique on some interval $[t_0,t]$. So, we take another $x_2$. Then we have two separate sets:
Set $A=left { s in [t_0,t] mid x_1 (s) neq x_2(s) right }$
Set $A^C=left { s in [t_0,t] mid x_1 (s) = x_2(s) right }$
We want to prove that $A^C=[t_0,t]$.
One thing that confuses me is that they proved that both of $A$ and $A^C$ are open sets. I won't write that proof here, but, they all make perfect sense. How can two complements be both open?
general-topology
general-topology
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
asked Jan 20 at 12:34
user25568user25568
1138
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
asked Jan 20 at 12:34
user25568user25568
1138
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
edited Jan 20 at 17:11
David C. Ullrich
61.1k43994
61.1k43994
asked Jan 20 at 12:34
user25568user25568
1138
asked Jan 20 at 12:34
user25568user25568
1138
asked Jan 20 at 12:34
user25568user25568
1138
1138
2
$begingroup$
A set can be both closed and open or neither, they're not each other's opposite
$endgroup$
– Alessandro Codenotti
Jan 20 at 12:36
1
$begingroup$
When one is empty, for instance. Or you can consider “disconnected” spaces, such as ${0,1}$.
$endgroup$
– Mindlack
Jan 20 at 12:37
add a comment |
2
$begingroup$
A set can be both closed and open or neither, they're not each other's opposite
$endgroup$
– Alessandro Codenotti
Jan 20 at 12:36
1
$begingroup$
When one is empty, for instance. Or you can consider “disconnected” spaces, such as ${0,1}$.
$endgroup$
– Mindlack
Jan 20 at 12:37
2
2
$begingroup$
A set can be both closed and open or neither, they're not each other's opposite
$endgroup$
– Alessandro Codenotti
Jan 20 at 12:36
$begingroup$
A set can be both closed and open or neither, they're not each other's opposite
$endgroup$
– Alessandro Codenotti
Jan 20 at 12:36
1
1
$begingroup$
When one is empty, for instance. Or you can consider “disconnected” spaces, such as ${0,1}$.
$endgroup$
– Mindlack
Jan 20 at 12:37
$begingroup$
When one is empty, for instance. Or you can consider “disconnected” spaces, such as ${0,1}$.
$endgroup$
– Mindlack
Jan 20 at 12:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Sets are not doors: they can be open and closed at the same time. And they can also be not open and not closed at the same time.
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
This appears to be a proof by contradiction. In such proofs the very idea is get something absurd to finish the proof. If $A$ is a subset of $[t_0,t]$ that is neither empty nor the whole interval then it is not possible for $A$ and $A^{c}$ to be compact and that is what makes the proof work.
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Sets are not doors: they can be open and closed at the same time. And they can also be not open and not closed at the same time.
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
Sets are not doors: they can be open and closed at the same time. And they can also be not open and not closed at the same time.
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
add a comment |
$begingroup$
Sets are not doors: they can be open and closed at the same time. And they can also be not open and not closed at the same time.
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
Sets are not doors: they can be open and closed at the same time. And they can also be not open and not closed at the same time.
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 12:37
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
add a comment |
add a comment |
$begingroup$
This appears to be a proof by contradiction. In such proofs the very idea is get something absurd to finish the proof. If $A$ is a subset of $[t_0,t]$ that is neither empty nor the whole interval then it is not possible for $A$ and $A^{c}$ to be compact and that is what makes the proof work.
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
$endgroup$
add a comment |
$begingroup$
This appears to be a proof by contradiction. In such proofs the very idea is get something absurd to finish the proof. If $A$ is a subset of $[t_0,t]$ that is neither empty nor the whole interval then it is not possible for $A$ and $A^{c}$ to be compact and that is what makes the proof work.
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
$endgroup$
add a comment |
$begingroup$
This appears to be a proof by contradiction. In such proofs the very idea is get something absurd to finish the proof. If $A$ is a subset of $[t_0,t]$ that is neither empty nor the whole interval then it is not possible for $A$ and $A^{c}$ to be compact and that is what makes the proof work.
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
$endgroup$
This appears to be a proof by contradiction. In such proofs the very idea is get something absurd to finish the proof. If $A$ is a subset of $[t_0,t]$ that is neither empty nor the whole interval then it is not possible for $A$ and $A^{c}$ to be compact and that is what makes the proof work.
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
answered Jan 20 at 12:37
Kavi Rama MurthyKavi Rama Murthy
64.4k42665
64.4k42665
add a comment |
add a comment |
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2
$begingroup$
A set can be both closed and open or neither, they're not each other's opposite
$endgroup$
– Alessandro Codenotti
Jan 20 at 12:36
1
$begingroup$
When one is empty, for instance. Or you can consider “disconnected” spaces, such as ${0,1}$.
$endgroup$
– Mindlack
Jan 20 at 12:37