Mixing
Is $DeclareMathOperatorintintint(Acup B)=int(A)cup int(B)$?
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I have posted a counterexample given in a solution below, but when I attempted the problem I did something different.
My Attempt
Take complements of both sides
$$
int(Acup B)stackrel?=int(A)cup int(B) \
big(int(Acup B)big)^cstackrel?=big(int(A)cup int(B)big)^c \
overline (Acup B)^cstackrel?=overline(A)^ccap overline(B)^c
$$
I got this step from this answer Prove that the closure of complement, is the complement of the interior
$$overline (A)^ccap (B)^cstackrel?=overline(A)^ccap overline(B)^c$$
My question
Could I prove from where I stopped that the two sides are not equal? I tried to sketch a ven diagram but both sides seem to give the same intersected set.
Counterexample
general-topology proof-writing
edited Jul 28 at 12:23
mvw
30.2k22250
asked Jul 28 at 11:44
john fowles
1,088817
add a comment |Â
up vote
0
down vote
favorite
I have posted a counterexample given in a solution below, but when I attempted the problem I did something different.
My Attempt
Take complements of both sides
$$
int(Acup B)stackrel?=int(A)cup int(B) \
big(int(Acup B)big)^cstackrel?=big(int(A)cup int(B)big)^c \
overline (Acup B)^cstackrel?=overline(A)^ccap overline(B)^c
$$
I got this step from this answer Prove that the closure of complement, is the complement of the interior
$$overline (A)^ccap (B)^cstackrel?=overline(A)^ccap overline(B)^c$$
My question
Could I prove from where I stopped that the two sides are not equal? I tried to sketch a ven diagram but both sides seem to give the same intersected set.
Counterexample
general-topology proof-writing
edited Jul 28 at 12:23
mvw
30.2k22250
asked Jul 28 at 11:44
john fowles
1,088817
From your example, you can get a counterexample to $overlineAcap B=overline Acapoverline B$ too.
– Lord Shark the Unknown
Jul 28 at 11:48
Let $A=(0,1)$ and $B=(1,2)$ thus $overlineAcap B=emptyset$ and $overline Acapoverline B=1$
– john fowles
Jul 28 at 12:01
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have posted a counterexample given in a solution below, but when I attempted the problem I did something different.
My Attempt
Take complements of both sides
$$
int(Acup B)stackrel?=int(A)cup int(B) \
big(int(Acup B)big)^cstackrel?=big(int(A)cup int(B)big)^c \
overline (Acup B)^cstackrel?=overline(A)^ccap overline(B)^c
$$
I got this step from this answer Prove that the closure of complement, is the complement of the interior
$$overline (A)^ccap (B)^cstackrel?=overline(A)^ccap overline(B)^c$$
My question
Could I prove from where I stopped that the two sides are not equal? I tried to sketch a ven diagram but both sides seem to give the same intersected set.
Counterexample
general-topology proof-writing
edited Jul 28 at 12:23
mvw
30.2k22250
asked Jul 28 at 11:44
john fowles
1,088817
I have posted a counterexample given in a solution below, but when I attempted the problem I did something different.
My Attempt
Take complements of both sides
$$
int(Acup B)stackrel?=int(A)cup int(B) \
big(int(Acup B)big)^cstackrel?=big(int(A)cup int(B)big)^c \
overline (Acup B)^cstackrel?=overline(A)^ccap overline(B)^c
$$
I got this step from this answer Prove that the closure of complement, is the complement of the interior
$$overline (A)^ccap (B)^cstackrel?=overline(A)^ccap overline(B)^c$$
My question
Could I prove from where I stopped that the two sides are not equal? I tried to sketch a ven diagram but both sides seem to give the same intersected set.
Counterexample
general-topology proof-writing
edited Jul 28 at 12:23
mvw
30.2k22250
asked Jul 28 at 11:44
john fowles
1,088817
edited Jul 28 at 12:23
mvw
30.2k22250
edited Jul 28 at 12:23
mvw
30.2k22250
edited Jul 28 at 12:23
mvw
30.2k22250
30.2k22250
asked Jul 28 at 11:44
john fowles
1,088817
asked Jul 28 at 11:44
john fowles
1,088817
asked Jul 28 at 11:44
john fowles
1,088817
1,088817
From your example, you can get a counterexample to $overlineAcap B=overline Acapoverline B$ too.
– Lord Shark the Unknown
Jul 28 at 11:48
Let $A=(0,1)$ and $B=(1,2)$ thus $overlineAcap B=emptyset$ and $overline Acapoverline B=1$
– john fowles
Jul 28 at 12:01
add a comment |Â
From your example, you can get a counterexample to $overlineAcap B=overline Acapoverline B$ too.
– Lord Shark the Unknown
Jul 28 at 11:48
Let $A=(0,1)$ and $B=(1,2)$ thus $overlineAcap B=emptyset$ and $overline Acapoverline B=1$
– john fowles
Jul 28 at 12:01
From your example, you can get a counterexample to $overlineAcap B=overline Acapoverline B$ too.
– Lord Shark the Unknown
Jul 28 at 11:48
From your example, you can get a counterexample to $overlineAcap B=overline Acapoverline B$ too.
– Lord Shark the Unknown
Jul 28 at 11:48
Let $A=(0,1)$ and $B=(1,2)$ thus $overlineAcap B=emptyset$ and $overline Acapoverline B=1$
– john fowles
Jul 28 at 12:01
Let $A=(0,1)$ and $B=(1,2)$ thus $overlineAcap B=emptyset$ and $overline Acapoverline B=1$
– john fowles
Jul 28 at 12:01
add a comment |Â
2 Answers
2
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oldest
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up vote
3
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accepted
Proposition. If the boundaries of A and B are disjoint, then
int A$cup$B = int A $cup$ int B.
answered Jul 28 at 11:57
William Elliot
5,0722414
add a comment |Â
up vote
4
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The correct answer to, "does $operatornameint(Acup B)$ equal $operatornameint(A)cupoperatornameint(B)$?" is:
Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.
So you have no hope of starting out from nothing and then prove that $operatornameint(Acup B) ne operatornameint(A)cupoperatornameint(B)$. That conclusion would be as wrong as it is to claim $operatornameint(Acup B) = operatornameint(A)cupoperatornameint(B)$.
For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false). You can't hope to do that by doing general manipulations starting from nothing, because that way -- if you do them correctly -- the only thing you will ever reach are things that always have such-and-such truth value.
Proving something not always true generally takes a concrete counterexample.
answered Jul 28 at 11:59
Henning Makholm
225k16290516
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Proposition. If the boundaries of A and B are disjoint, then
int A$cup$B = int A $cup$ int B.
answered Jul 28 at 11:57
William Elliot
5,0722414
add a comment |Â
up vote
3
down vote
accepted
Proposition. If the boundaries of A and B are disjoint, then
int A$cup$B = int A $cup$ int B.
answered Jul 28 at 11:57
William Elliot
5,0722414
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Proposition. If the boundaries of A and B are disjoint, then
int A$cup$B = int A $cup$ int B.
answered Jul 28 at 11:57
William Elliot
5,0722414
Proposition. If the boundaries of A and B are disjoint, then
int A$cup$B = int A $cup$ int B.
answered Jul 28 at 11:57
William Elliot
5,0722414
answered Jul 28 at 11:57
William Elliot
5,0722414
answered Jul 28 at 11:57
William Elliot
5,0722414
answered Jul 28 at 11:57
William Elliot
5,0722414
5,0722414
add a comment |Â
add a comment |Â
up vote
4
down vote
The correct answer to, "does $operatornameint(Acup B)$ equal $operatornameint(A)cupoperatornameint(B)$?" is:
Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.
So you have no hope of starting out from nothing and then prove that $operatornameint(Acup B) ne operatornameint(A)cupoperatornameint(B)$. That conclusion would be as wrong as it is to claim $operatornameint(Acup B) = operatornameint(A)cupoperatornameint(B)$.
For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false). You can't hope to do that by doing general manipulations starting from nothing, because that way -- if you do them correctly -- the only thing you will ever reach are things that always have such-and-such truth value.
Proving something not always true generally takes a concrete counterexample.
answered Jul 28 at 11:59
Henning Makholm
225k16290516
add a comment |Â
up vote
4
down vote
The correct answer to, "does $operatornameint(Acup B)$ equal $operatornameint(A)cupoperatornameint(B)$?" is:
Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.
So you have no hope of starting out from nothing and then prove that $operatornameint(Acup B) ne operatornameint(A)cupoperatornameint(B)$. That conclusion would be as wrong as it is to claim $operatornameint(Acup B) = operatornameint(A)cupoperatornameint(B)$.
For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false). You can't hope to do that by doing general manipulations starting from nothing, because that way -- if you do them correctly -- the only thing you will ever reach are things that always have such-and-such truth value.
Proving something not always true generally takes a concrete counterexample.
answered Jul 28 at 11:59
Henning Makholm
225k16290516
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The correct answer to, "does $operatornameint(Acup B)$ equal $operatornameint(A)cupoperatornameint(B)$?" is:
Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.
So you have no hope of starting out from nothing and then prove that $operatornameint(Acup B) ne operatornameint(A)cupoperatornameint(B)$. That conclusion would be as wrong as it is to claim $operatornameint(Acup B) = operatornameint(A)cupoperatornameint(B)$.
For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false). You can't hope to do that by doing general manipulations starting from nothing, because that way -- if you do them correctly -- the only thing you will ever reach are things that always have such-and-such truth value.
Proving something not always true generally takes a concrete counterexample.
answered Jul 28 at 11:59
Henning Makholm
225k16290516
The correct answer to, "does $operatornameint(Acup B)$ equal $operatornameint(A)cupoperatornameint(B)$?" is:
Perhaps yes, perhaps no. That depends on which sets $A$ and $B$ are.
So you have no hope of starting out from nothing and then prove that $operatornameint(Acup B) ne operatornameint(A)cupoperatornameint(B)$. That conclusion would be as wrong as it is to claim $operatornameint(Acup B) = operatornameint(A)cupoperatornameint(B)$.
For a claim (with parameters) that is sometimes true and sometimes false, the best you can do is to prove that it is not always true (and subsequently that it is not always false). You can't hope to do that by doing general manipulations starting from nothing, because that way -- if you do them correctly -- the only thing you will ever reach are things that always have such-and-such truth value.
Proving something not always true generally takes a concrete counterexample.
answered Jul 28 at 11:59
Henning Makholm
225k16290516
answered Jul 28 at 11:59
Henning Makholm
225k16290516
answered Jul 28 at 11:59
Henning Makholm
225k16290516
answered Jul 28 at 11:59
Henning Makholm
225k16290516
225k16290516
add a comment |Â
add a comment |Â
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From your example, you can get a counterexample to $overlineAcap B=overline Acapoverline B$ too.
– Lord Shark the Unknown
Jul 28 at 11:48
Let $A=(0,1)$ and $B=(1,2)$ thus $overlineAcap B=emptyset$ and $overline Acapoverline B=1$
– john fowles
Jul 28 at 12:01