A counterexample to “a associates b”, then “a strongly associates b”
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In this article, the author gives the following counterexample for "a associates b" but not "a strongly associates b" at the end of the first page.
However, in $[1,2]$, the values of both $x,y$ are zero, so there is no need for choosing $c(t) = 3- 2t$ in that interval, i.e choosing $c(t) = 1 $ in $[1,2]$ is also a possible choice, and in the latter case, $c$ is a unit element of $C[0,3]$ which invalidates this "counterexample".
Considering the fact that this article is read by many mathematicians and published, what is wrong in my view that this example is still a counterexample ?
abstract-algebra ring-theory
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
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add a comment |
$begingroup$
In this article, the author gives the following counterexample for "a associates b" but not "a strongly associates b" at the end of the first page.
However, in $[1,2]$, the values of both $x,y$ are zero, so there is no need for choosing $c(t) = 3- 2t$ in that interval, i.e choosing $c(t) = 1 $ in $[1,2]$ is also a possible choice, and in the latter case, $c$ is a unit element of $C[0,3]$ which invalidates this "counterexample".
Considering the fact that this article is read by many mathematicians and published, what is wrong in my view that this example is still a counterexample ?
abstract-algebra ring-theory
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
$endgroup$
add a comment |
$begingroup$
In this article, the author gives the following counterexample for "a associates b" but not "a strongly associates b" at the end of the first page.
However, in $[1,2]$, the values of both $x,y$ are zero, so there is no need for choosing $c(t) = 3- 2t$ in that interval, i.e choosing $c(t) = 1 $ in $[1,2]$ is also a possible choice, and in the latter case, $c$ is a unit element of $C[0,3]$ which invalidates this "counterexample".
Considering the fact that this article is read by many mathematicians and published, what is wrong in my view that this example is still a counterexample ?
abstract-algebra ring-theory
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
$endgroup$
In this article, the author gives the following counterexample for "a associates b" but not "a strongly associates b" at the end of the first page.
However, in $[1,2]$, the values of both $x,y$ are zero, so there is no need for choosing $c(t) = 3- 2t$ in that interval, i.e choosing $c(t) = 1 $ in $[1,2]$ is also a possible choice, and in the latter case, $c$ is a unit element of $C[0,3]$ which invalidates this "counterexample".
Considering the fact that this article is read by many mathematicians and published, what is wrong in my view that this example is still a counterexample ?
abstract-algebra ring-theory
abstract-algebra ring-theory
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
asked Dec 18 '18 at 5:58
onurcanbektasonurcanbektas
3,36011036
3,36011036
add a comment |
add a comment |
1 Answer
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Your proposed definition of $c$ is not continuous, and so is not an element of $C[0,3]$. The point is that $c$ needs to be $1$ on $[0,1]$ and $-1$ on $[2,3]$, so in order to be continuous it would need to pass through $0$ in between, and so it cannot be a unit.
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your proposed definition of $c$ is not continuous, and so is not an element of $C[0,3]$. The point is that $c$ needs to be $1$ on $[0,1]$ and $-1$ on $[2,3]$, so in order to be continuous it would need to pass through $0$ in between, and so it cannot be a unit.
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
Your proposed definition of $c$ is not continuous, and so is not an element of $C[0,3]$. The point is that $c$ needs to be $1$ on $[0,1]$ and $-1$ on $[2,3]$, so in order to be continuous it would need to pass through $0$ in between, and so it cannot be a unit.
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
$endgroup$
add a comment |
$begingroup$
Your proposed definition of $c$ is not continuous, and so is not an element of $C[0,3]$. The point is that $c$ needs to be $1$ on $[0,1]$ and $-1$ on $[2,3]$, so in order to be continuous it would need to pass through $0$ in between, and so it cannot be a unit.
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
$endgroup$
Your proposed definition of $c$ is not continuous, and so is not an element of $C[0,3]$. The point is that $c$ needs to be $1$ on $[0,1]$ and $-1$ on $[2,3]$, so in order to be continuous it would need to pass through $0$ in between, and so it cannot be a unit.
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
answered Dec 18 '18 at 6:03
Eric WofseyEric Wofsey
183k13209337
183k13209337
add a comment |
add a comment |
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