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Prove or disprove that the set of continuous mappings $phicolon [0;1]to [0;1]$ such that […] forms a group.
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The full question: Prove or disprove that the set of all continuous relations mappings $ phicolon [0;1]to [0;1]$, for which $phi(0)=0$, $phi(1)=1$ and $x<y implies phi(x)< phi(y)$ forms a group with respect to the operation of superposition/composition.
I was able to prove closure by noting that the intersection of their domains is simply the interval $[0;1]$ itself thus the composition will also be a relation mapping from $[0;1]$ to $[0;1]$.
At this point, I decided to prove associativity. I first considered this well-known proof:
$$hcirc(gcirc f)(x)=h(gcirc f(x))= h(g(f(x))=hcirc g(f(x))=(hcirc g)circ f(x)$$
However, this is a proof for function composition whereas the question is about relations mappings. So, since the question asked about relations mappings and not just functions, is this proof of associativity still valid? After all, it seems to me that all the continuous increasing relations mappings from $[0;1]$ to $[0;1]$ are indeed functions. (I'm guessing this statement would require it's own proof, which i'm trying to draft as well.)
Stuck on associativity, I decided to see if the other group axioms would be easier to work with. I verified than an identity function $i(x)=x$ exists, satisfying the conditions of the problem. Then I had more trouble with proving that the inverses will also be continious and within the interval so I'm inclined to conclude that the set is not a group. Any hints on how to proceed with this proof or disproof?
abstract-algebra group-theory relations
asked 2 days ago
E.Nole
1716
|
show 6 more comments
up vote
0
down vote
favorite
The full question: Prove or disprove that the set of all continuous relations mappings $ phicolon [0;1]to [0;1]$, for which $phi(0)=0$, $phi(1)=1$ and $x<y implies phi(x)< phi(y)$ forms a group with respect to the operation of superposition/composition.
I was able to prove closure by noting that the intersection of their domains is simply the interval $[0;1]$ itself thus the composition will also be a relation mapping from $[0;1]$ to $[0;1]$.
At this point, I decided to prove associativity. I first considered this well-known proof:
$$hcirc(gcirc f)(x)=h(gcirc f(x))= h(g(f(x))=hcirc g(f(x))=(hcirc g)circ f(x)$$
However, this is a proof for function composition whereas the question is about relations mappings. So, since the question asked about relations mappings and not just functions, is this proof of associativity still valid? After all, it seems to me that all the continuous increasing relations mappings from $[0;1]$ to $[0;1]$ are indeed functions. (I'm guessing this statement would require it's own proof, which i'm trying to draft as well.)
Stuck on associativity, I decided to see if the other group axioms would be easier to work with. I verified than an identity function $i(x)=x$ exists, satisfying the conditions of the problem. Then I had more trouble with proving that the inverses will also be continious and within the interval so I'm inclined to conclude that the set is not a group. Any hints on how to proceed with this proof or disproof?
abstract-algebra group-theory relations
asked 2 days ago
E.Nole
1716
1
What do you mean by "continuous relation"? What does $phi(x)$ mean if $phi$ is a relation?
– Eric Wofsey
2 days ago
I'm not sure I understand your question. (It seems there is some subtext?) Well, a relation is any set of ordered pairs. A continuous relation is a relation which is defined for all points in a given interval.
– E.Nole
2 days ago
OK, a relation is a set of ordered pairs. What does $phi(0)=0$ mean, then, if $phi$ is a set of ordered pairs? (The subtext is that "continuous relation" is not a standard term at all, nor is it standard to use notation like $phi(x)$ when talking about relations.)
– Eric Wofsey
2 days ago
And, what does it mean for a relation to be "defined" for a point? Maybe $phi$ is defined for $x$ if there exists $y$ such that $(x,y)inphi$?
– Eric Wofsey
2 days ago
The problem was translated from a Russian text so it's possible some details were missed. However it should be more or less clear what is being asked. This is how I understand it: Let G be the set of relations $ phi_n$ from $[0;1]$ to $[0;1]$ satisfying the conditions described in the question. The question is whether or not G and the operation of composition forms a group. Does this help clarify?
– E.Nole
2 days ago
|
show 6 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The full question: Prove or disprove that the set of all continuous relations mappings $ phicolon [0;1]to [0;1]$, for which $phi(0)=0$, $phi(1)=1$ and $x<y implies phi(x)< phi(y)$ forms a group with respect to the operation of superposition/composition.
I was able to prove closure by noting that the intersection of their domains is simply the interval $[0;1]$ itself thus the composition will also be a relation mapping from $[0;1]$ to $[0;1]$.
At this point, I decided to prove associativity. I first considered this well-known proof:
$$hcirc(gcirc f)(x)=h(gcirc f(x))= h(g(f(x))=hcirc g(f(x))=(hcirc g)circ f(x)$$
However, this is a proof for function composition whereas the question is about relations mappings. So, since the question asked about relations mappings and not just functions, is this proof of associativity still valid? After all, it seems to me that all the continuous increasing relations mappings from $[0;1]$ to $[0;1]$ are indeed functions. (I'm guessing this statement would require it's own proof, which i'm trying to draft as well.)
Stuck on associativity, I decided to see if the other group axioms would be easier to work with. I verified than an identity function $i(x)=x$ exists, satisfying the conditions of the problem. Then I had more trouble with proving that the inverses will also be continious and within the interval so I'm inclined to conclude that the set is not a group. Any hints on how to proceed with this proof or disproof?
abstract-algebra group-theory relations
asked 2 days ago
E.Nole
1716
The full question: Prove or disprove that the set of all continuous relations mappings $ phicolon [0;1]to [0;1]$, for which $phi(0)=0$, $phi(1)=1$ and $x<y implies phi(x)< phi(y)$ forms a group with respect to the operation of superposition/composition.
I was able to prove closure by noting that the intersection of their domains is simply the interval $[0;1]$ itself thus the composition will also be a relation mapping from $[0;1]$ to $[0;1]$.
At this point, I decided to prove associativity. I first considered this well-known proof:
$$hcirc(gcirc f)(x)=h(gcirc f(x))= h(g(f(x))=hcirc g(f(x))=(hcirc g)circ f(x)$$
However, this is a proof for function composition whereas the question is about relations mappings. So, since the question asked about relations mappings and not just functions, is this proof of associativity still valid? After all, it seems to me that all the continuous increasing relations mappings from $[0;1]$ to $[0;1]$ are indeed functions. (I'm guessing this statement would require it's own proof, which i'm trying to draft as well.)
Stuck on associativity, I decided to see if the other group axioms would be easier to work with. I verified than an identity function $i(x)=x$ exists, satisfying the conditions of the problem. Then I had more trouble with proving that the inverses will also be continious and within the interval so I'm inclined to conclude that the set is not a group. Any hints on how to proceed with this proof or disproof?
abstract-algebra group-theory relations
abstract-algebra group-theory relations
asked 2 days ago
E.Nole
1716
asked 2 days ago
E.Nole
1716
edited 2 days ago
asked 2 days ago
E.Nole
1716
asked 2 days ago
E.Nole
1716
asked 2 days ago
E.Nole
1716
1716
1
What do you mean by "continuous relation"? What does $phi(x)$ mean if $phi$ is a relation?
– Eric Wofsey
2 days ago
I'm not sure I understand your question. (It seems there is some subtext?) Well, a relation is any set of ordered pairs. A continuous relation is a relation which is defined for all points in a given interval.
– E.Nole
2 days ago
OK, a relation is a set of ordered pairs. What does $phi(0)=0$ mean, then, if $phi$ is a set of ordered pairs? (The subtext is that "continuous relation" is not a standard term at all, nor is it standard to use notation like $phi(x)$ when talking about relations.)
– Eric Wofsey
2 days ago
And, what does it mean for a relation to be "defined" for a point? Maybe $phi$ is defined for $x$ if there exists $y$ such that $(x,y)inphi$?
– Eric Wofsey
2 days ago
The problem was translated from a Russian text so it's possible some details were missed. However it should be more or less clear what is being asked. This is how I understand it: Let G be the set of relations $ phi_n$ from $[0;1]$ to $[0;1]$ satisfying the conditions described in the question. The question is whether or not G and the operation of composition forms a group. Does this help clarify?
– E.Nole
2 days ago
|
show 6 more comments
1
What do you mean by "continuous relation"? What does $phi(x)$ mean if $phi$ is a relation?
– Eric Wofsey
2 days ago
I'm not sure I understand your question. (It seems there is some subtext?) Well, a relation is any set of ordered pairs. A continuous relation is a relation which is defined for all points in a given interval.
– E.Nole
2 days ago
OK, a relation is a set of ordered pairs. What does $phi(0)=0$ mean, then, if $phi$ is a set of ordered pairs? (The subtext is that "continuous relation" is not a standard term at all, nor is it standard to use notation like $phi(x)$ when talking about relations.)
– Eric Wofsey
2 days ago
And, what does it mean for a relation to be "defined" for a point? Maybe $phi$ is defined for $x$ if there exists $y$ such that $(x,y)inphi$?
– Eric Wofsey
2 days ago
The problem was translated from a Russian text so it's possible some details were missed. However it should be more or less clear what is being asked. This is how I understand it: Let G be the set of relations $ phi_n$ from $[0;1]$ to $[0;1]$ satisfying the conditions described in the question. The question is whether or not G and the operation of composition forms a group. Does this help clarify?
– E.Nole
2 days ago
1
1
What do you mean by "continuous relation"? What does $phi(x)$ mean if $phi$ is a relation?
– Eric Wofsey
2 days ago
What do you mean by "continuous relation"? What does $phi(x)$ mean if $phi$ is a relation?
– Eric Wofsey
2 days ago
I'm not sure I understand your question. (It seems there is some subtext?) Well, a relation is any set of ordered pairs. A continuous relation is a relation which is defined for all points in a given interval.
– E.Nole
2 days ago
I'm not sure I understand your question. (It seems there is some subtext?) Well, a relation is any set of ordered pairs. A continuous relation is a relation which is defined for all points in a given interval.
– E.Nole
2 days ago
OK, a relation is a set of ordered pairs. What does $phi(0)=0$ mean, then, if $phi$ is a set of ordered pairs? (The subtext is that "continuous relation" is not a standard term at all, nor is it standard to use notation like $phi(x)$ when talking about relations.)
– Eric Wofsey
2 days ago
OK, a relation is a set of ordered pairs. What does $phi(0)=0$ mean, then, if $phi$ is a set of ordered pairs? (The subtext is that "continuous relation" is not a standard term at all, nor is it standard to use notation like $phi(x)$ when talking about relations.)
– Eric Wofsey
2 days ago
And, what does it mean for a relation to be "defined" for a point? Maybe $phi$ is defined for $x$ if there exists $y$ such that $(x,y)inphi$?
– Eric Wofsey
2 days ago
And, what does it mean for a relation to be "defined" for a point? Maybe $phi$ is defined for $x$ if there exists $y$ such that $(x,y)inphi$?
– Eric Wofsey
2 days ago
The problem was translated from a Russian text so it's possible some details were missed. However it should be more or less clear what is being asked. This is how I understand it: Let G be the set of relations $ phi_n$ from $[0;1]$ to $[0;1]$ satisfying the conditions described in the question. The question is whether or not G and the operation of composition forms a group. Does this help clarify?
– E.Nole
2 days ago
The problem was translated from a Russian text so it's possible some details were missed. However it should be more or less clear what is being asked. This is how I understand it: Let G be the set of relations $ phi_n$ from $[0;1]$ to $[0;1]$ satisfying the conditions described in the question. The question is whether or not G and the operation of composition forms a group. Does this help clarify?
– E.Nole
2 days ago
|
show 6 more comments
1 Answer
1
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up vote
1
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$require{AMScd}$
So first of all the associativity follows from the general fact about functions: if we have
$$begin{CD}
A @>{f}>> B @>{g}>> C @>{h}>> D
end{CD} $$
then $hcirc(gcirc f)=(hcirc g)circ f$ regardless of what those functions are. This can be checked by evaluating each side at $xin A$ as you've done yourself.
Now the set of all bijections (not necessarily continuous) from a set to itself is a group together with $circ$ as a group multiplication. Denote this group by $mathcal{F}(X)$.
So now it is enough to show that the set you are looking at is a subgroup of $mathcal{F}(X)$.
(1) So first of all if $f:[0,1]to[0,1]$ is such that $f(0)=0$, $f(1)=1$ and $f$ is continuous and monotonic then $f$ is a bijection. This follows from the intermediate value theorem.
Denote by $mathcal{M}([0,1])$ the set of all those functions.
(2) The set $mathcal{M}([0,1])$ is closed under $circ$. This you claim you've shown. The proof is quite simple though. If $x<y$ then $f(x)<f(y)$ and thus so is $g(f(x))<g(f(y))$. Meaning $gcirc finmathcal{M}([0,1])$ if both $f,ginmathcal{M}([0,1])$.
(3) If $fin mathcal{M}([0,1])$ then $f^{-1}inmathcal{M}([0,1])$. So obviously $f^{-1}(0)=0$ and $f^{-1}(1)=1$. Now we will show that $f^{-1}$ is monotonic. Indeed, assume that $x<y$. Then $x=f(a)$ and $y=f(b)$ since $f$ is bijective. But since $f$ is monotonic then case when $a=b$ or $a>b$ are impossible. It follows that $a<b$ and thus $f^{-1}(x)<f^{-1}(y)$.
Finally $f^{-1}$ is continuous simply because $[0,1]$ is compact and so $f$ is closed.
answered 2 days ago
freakish
10.4k1526
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$require{AMScd}$
So first of all the associativity follows from the general fact about functions: if we have
$$begin{CD}
A @>{f}>> B @>{g}>> C @>{h}>> D
end{CD} $$
then $hcirc(gcirc f)=(hcirc g)circ f$ regardless of what those functions are. This can be checked by evaluating each side at $xin A$ as you've done yourself.
Now the set of all bijections (not necessarily continuous) from a set to itself is a group together with $circ$ as a group multiplication. Denote this group by $mathcal{F}(X)$.
So now it is enough to show that the set you are looking at is a subgroup of $mathcal{F}(X)$.
(1) So first of all if $f:[0,1]to[0,1]$ is such that $f(0)=0$, $f(1)=1$ and $f$ is continuous and monotonic then $f$ is a bijection. This follows from the intermediate value theorem.
Denote by $mathcal{M}([0,1])$ the set of all those functions.
(2) The set $mathcal{M}([0,1])$ is closed under $circ$. This you claim you've shown. The proof is quite simple though. If $x<y$ then $f(x)<f(y)$ and thus so is $g(f(x))<g(f(y))$. Meaning $gcirc finmathcal{M}([0,1])$ if both $f,ginmathcal{M}([0,1])$.
(3) If $fin mathcal{M}([0,1])$ then $f^{-1}inmathcal{M}([0,1])$. So obviously $f^{-1}(0)=0$ and $f^{-1}(1)=1$. Now we will show that $f^{-1}$ is monotonic. Indeed, assume that $x<y$. Then $x=f(a)$ and $y=f(b)$ since $f$ is bijective. But since $f$ is monotonic then case when $a=b$ or $a>b$ are impossible. It follows that $a<b$ and thus $f^{-1}(x)<f^{-1}(y)$.
Finally $f^{-1}$ is continuous simply because $[0,1]$ is compact and so $f$ is closed.
answered 2 days ago
freakish
10.4k1526
add a comment |
up vote
1
down vote
$require{AMScd}$
So first of all the associativity follows from the general fact about functions: if we have
$$begin{CD}
A @>{f}>> B @>{g}>> C @>{h}>> D
end{CD} $$
then $hcirc(gcirc f)=(hcirc g)circ f$ regardless of what those functions are. This can be checked by evaluating each side at $xin A$ as you've done yourself.
Now the set of all bijections (not necessarily continuous) from a set to itself is a group together with $circ$ as a group multiplication. Denote this group by $mathcal{F}(X)$.
So now it is enough to show that the set you are looking at is a subgroup of $mathcal{F}(X)$.
(1) So first of all if $f:[0,1]to[0,1]$ is such that $f(0)=0$, $f(1)=1$ and $f$ is continuous and monotonic then $f$ is a bijection. This follows from the intermediate value theorem.
Denote by $mathcal{M}([0,1])$ the set of all those functions.
(2) The set $mathcal{M}([0,1])$ is closed under $circ$. This you claim you've shown. The proof is quite simple though. If $x<y$ then $f(x)<f(y)$ and thus so is $g(f(x))<g(f(y))$. Meaning $gcirc finmathcal{M}([0,1])$ if both $f,ginmathcal{M}([0,1])$.
(3) If $fin mathcal{M}([0,1])$ then $f^{-1}inmathcal{M}([0,1])$. So obviously $f^{-1}(0)=0$ and $f^{-1}(1)=1$. Now we will show that $f^{-1}$ is monotonic. Indeed, assume that $x<y$. Then $x=f(a)$ and $y=f(b)$ since $f$ is bijective. But since $f$ is monotonic then case when $a=b$ or $a>b$ are impossible. It follows that $a<b$ and thus $f^{-1}(x)<f^{-1}(y)$.
Finally $f^{-1}$ is continuous simply because $[0,1]$ is compact and so $f$ is closed.
answered 2 days ago
freakish
10.4k1526
add a comment |
up vote
1
down vote
up vote
1
down vote
$require{AMScd}$
So first of all the associativity follows from the general fact about functions: if we have
$$begin{CD}
A @>{f}>> B @>{g}>> C @>{h}>> D
end{CD} $$
then $hcirc(gcirc f)=(hcirc g)circ f$ regardless of what those functions are. This can be checked by evaluating each side at $xin A$ as you've done yourself.
Now the set of all bijections (not necessarily continuous) from a set to itself is a group together with $circ$ as a group multiplication. Denote this group by $mathcal{F}(X)$.
So now it is enough to show that the set you are looking at is a subgroup of $mathcal{F}(X)$.
(1) So first of all if $f:[0,1]to[0,1]$ is such that $f(0)=0$, $f(1)=1$ and $f$ is continuous and monotonic then $f$ is a bijection. This follows from the intermediate value theorem.
Denote by $mathcal{M}([0,1])$ the set of all those functions.
(2) The set $mathcal{M}([0,1])$ is closed under $circ$. This you claim you've shown. The proof is quite simple though. If $x<y$ then $f(x)<f(y)$ and thus so is $g(f(x))<g(f(y))$. Meaning $gcirc finmathcal{M}([0,1])$ if both $f,ginmathcal{M}([0,1])$.
(3) If $fin mathcal{M}([0,1])$ then $f^{-1}inmathcal{M}([0,1])$. So obviously $f^{-1}(0)=0$ and $f^{-1}(1)=1$. Now we will show that $f^{-1}$ is monotonic. Indeed, assume that $x<y$. Then $x=f(a)$ and $y=f(b)$ since $f$ is bijective. But since $f$ is monotonic then case when $a=b$ or $a>b$ are impossible. It follows that $a<b$ and thus $f^{-1}(x)<f^{-1}(y)$.
Finally $f^{-1}$ is continuous simply because $[0,1]$ is compact and so $f$ is closed.
answered 2 days ago
freakish
10.4k1526
$require{AMScd}$
So first of all the associativity follows from the general fact about functions: if we have
$$begin{CD}
A @>{f}>> B @>{g}>> C @>{h}>> D
end{CD} $$
then $hcirc(gcirc f)=(hcirc g)circ f$ regardless of what those functions are. This can be checked by evaluating each side at $xin A$ as you've done yourself.
Now the set of all bijections (not necessarily continuous) from a set to itself is a group together with $circ$ as a group multiplication. Denote this group by $mathcal{F}(X)$.
So now it is enough to show that the set you are looking at is a subgroup of $mathcal{F}(X)$.
(1) So first of all if $f:[0,1]to[0,1]$ is such that $f(0)=0$, $f(1)=1$ and $f$ is continuous and monotonic then $f$ is a bijection. This follows from the intermediate value theorem.
Denote by $mathcal{M}([0,1])$ the set of all those functions.
(2) The set $mathcal{M}([0,1])$ is closed under $circ$. This you claim you've shown. The proof is quite simple though. If $x<y$ then $f(x)<f(y)$ and thus so is $g(f(x))<g(f(y))$. Meaning $gcirc finmathcal{M}([0,1])$ if both $f,ginmathcal{M}([0,1])$.
(3) If $fin mathcal{M}([0,1])$ then $f^{-1}inmathcal{M}([0,1])$. So obviously $f^{-1}(0)=0$ and $f^{-1}(1)=1$. Now we will show that $f^{-1}$ is monotonic. Indeed, assume that $x<y$. Then $x=f(a)$ and $y=f(b)$ since $f$ is bijective. But since $f$ is monotonic then case when $a=b$ or $a>b$ are impossible. It follows that $a<b$ and thus $f^{-1}(x)<f^{-1}(y)$.
Finally $f^{-1}$ is continuous simply because $[0,1]$ is compact and so $f$ is closed.
answered 2 days ago
freakish
10.4k1526
edited 2 days ago
answered 2 days ago
freakish
10.4k1526
answered 2 days ago
freakish
10.4k1526
answered 2 days ago
freakish
10.4k1526
10.4k1526
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1
What do you mean by "continuous relation"? What does $phi(x)$ mean if $phi$ is a relation?
– Eric Wofsey
2 days ago
I'm not sure I understand your question. (It seems there is some subtext?) Well, a relation is any set of ordered pairs. A continuous relation is a relation which is defined for all points in a given interval.
– E.Nole
2 days ago
OK, a relation is a set of ordered pairs. What does $phi(0)=0$ mean, then, if $phi$ is a set of ordered pairs? (The subtext is that "continuous relation" is not a standard term at all, nor is it standard to use notation like $phi(x)$ when talking about relations.)
– Eric Wofsey
2 days ago
And, what does it mean for a relation to be "defined" for a point? Maybe $phi$ is defined for $x$ if there exists $y$ such that $(x,y)inphi$?
– Eric Wofsey
2 days ago
The problem was translated from a Russian text so it's possible some details were missed. However it should be more or less clear what is being asked. This is how I understand it: Let G be the set of relations $ phi_n$ from $[0;1]$ to $[0;1]$ satisfying the conditions described in the question. The question is whether or not G and the operation of composition forms a group. Does this help clarify?
– E.Nole
2 days ago