Mixing
$ (a,b), (c,d) in mathbb{R} times mathbb{R} $ let us define a relation by $(a,b) sim (c,d)$ if and only if $...
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Question:
For $ (a,b), (c,d) in mathbb{R} times mathbb{R} $ let us define a relation by $(a,b) sim (c,d)$ if and only if $ a + 2d = c+2b$
Is this an equivalence relation on $mathbb{R} times mathbb{R}$?
My attempt:
Reflexive?
Notice that $ forall (a,b) in mathbb{R} times mathbb{R}, a + 2b = a + 2b implies ((a,b),(a,b)) in R$.
Hence the relation is reflexive.
Symmetric?
If $ ((a,b), (c,d)) in R implies a + 2d = c + 2b implies c + 2b = a + 2d implies ((c,d), (a,b)) in R $.
Hence the relation is symmetric.
Transitive?
If $ ((a,b), (c,d)) in R$ and $ ((c,d), (e,f)) in R implies a + 2d = c+2b $ and $ c + 2f = e + 2d implies a + 2d + c + 2f = c+2b + e + 2d implies a + 2f = 2b + e implies a + 2f = e + 2b implies ((a,b),(e,f)) in R$
Hence the relation is transitive.
Therefore, the relation is an equivalence relation.
I am not quite sure if I have proved it correctly and if my approach is correct.
proof-verification relations equivalence-relations
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Question:
For $ (a,b), (c,d) in mathbb{R} times mathbb{R} $ let us define a relation by $(a,b) sim (c,d)$ if and only if $ a + 2d = c+2b$
Is this an equivalence relation on $mathbb{R} times mathbb{R}$?
My attempt:
Reflexive?
Notice that $ forall (a,b) in mathbb{R} times mathbb{R}, a + 2b = a + 2b implies ((a,b),(a,b)) in R$.
Hence the relation is reflexive.
Symmetric?
If $ ((a,b), (c,d)) in R implies a + 2d = c + 2b implies c + 2b = a + 2d implies ((c,d), (a,b)) in R $.
Hence the relation is symmetric.
Transitive?
If $ ((a,b), (c,d)) in R$ and $ ((c,d), (e,f)) in R implies a + 2d = c+2b $ and $ c + 2f = e + 2d implies a + 2d + c + 2f = c+2b + e + 2d implies a + 2f = 2b + e implies a + 2f = e + 2b implies ((a,b),(e,f)) in R$
Hence the relation is transitive.
Therefore, the relation is an equivalence relation.
I am not quite sure if I have proved it correctly and if my approach is correct.
proof-verification relations equivalence-relations
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Question:
For $ (a,b), (c,d) in mathbb{R} times mathbb{R} $ let us define a relation by $(a,b) sim (c,d)$ if and only if $ a + 2d = c+2b$
Is this an equivalence relation on $mathbb{R} times mathbb{R}$?
My attempt:
Reflexive?
Notice that $ forall (a,b) in mathbb{R} times mathbb{R}, a + 2b = a + 2b implies ((a,b),(a,b)) in R$.
Hence the relation is reflexive.
Symmetric?
If $ ((a,b), (c,d)) in R implies a + 2d = c + 2b implies c + 2b = a + 2d implies ((c,d), (a,b)) in R $.
Hence the relation is symmetric.
Transitive?
If $ ((a,b), (c,d)) in R$ and $ ((c,d), (e,f)) in R implies a + 2d = c+2b $ and $ c + 2f = e + 2d implies a + 2d + c + 2f = c+2b + e + 2d implies a + 2f = 2b + e implies a + 2f = e + 2b implies ((a,b),(e,f)) in R$
Hence the relation is transitive.
Therefore, the relation is an equivalence relation.
I am not quite sure if I have proved it correctly and if my approach is correct.
proof-verification relations equivalence-relations
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
$endgroup$
Question:
For $ (a,b), (c,d) in mathbb{R} times mathbb{R} $ let us define a relation by $(a,b) sim (c,d)$ if and only if $ a + 2d = c+2b$
Is this an equivalence relation on $mathbb{R} times mathbb{R}$?
My attempt:
Reflexive?
Notice that $ forall (a,b) in mathbb{R} times mathbb{R}, a + 2b = a + 2b implies ((a,b),(a,b)) in R$.
Hence the relation is reflexive.
Symmetric?
If $ ((a,b), (c,d)) in R implies a + 2d = c + 2b implies c + 2b = a + 2d implies ((c,d), (a,b)) in R $.
Hence the relation is symmetric.
Transitive?
If $ ((a,b), (c,d)) in R$ and $ ((c,d), (e,f)) in R implies a + 2d = c+2b $ and $ c + 2f = e + 2d implies a + 2d + c + 2f = c+2b + e + 2d implies a + 2f = 2b + e implies a + 2f = e + 2b implies ((a,b),(e,f)) in R$
Hence the relation is transitive.
Therefore, the relation is an equivalence relation.
I am not quite sure if I have proved it correctly and if my approach is correct.
proof-verification relations equivalence-relations
proof-verification relations equivalence-relations
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
edited Aug 11 '17 at 3:26
zipirovich
11.3k11731
11.3k11731
asked Aug 11 '17 at 3:05
user444945
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3 Answers
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Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
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That's a correct direct approach. More generally note that $,(a,b)sim (c,d)iff f(a,b) = f(c,d),$ for $,f(u,v) = u-2v,$ so we can apply the following very general kernel criterion.
It is straightforward to prove relations of form $rm, xsim y {overset{ def}{color{#c00}iff}} f(x) = f(y), $ are equivalence relations.
More generally, suppose $rm usim v smash[t]{overset{ def}{color{#c00}iff}}, f(u) approx f(v) $ for a function $rm,f,$ and equivalence relation $,approx., $ Then the equivalence relation $rmcolor{#0a0}{properties (E)},$ of $,approx,$ transport (pullback) to $,sim,$ along $rm,f$ as follows
reflexive $rmquad color{#0a0}{overset{(E)}Rightarrow}, f(v) approx f(v):color{#c00}Rightarrow:vsim v$
symmetric $rm, usim v:color{#c00}Rightarrow f(u) approx f(v):color{#0a0}{overset{(E)}Rightarrow}:f(v)approx f(u):color{#c00}Rightarrow:vsim u$
transitive $rm usim v,, vsim w:color{#c00}Rightarrow: f(u)approx f(v),,f(v)approx f(w):color{#0a0}{overset{(E)}Rightarrow}:f(u)approx f(w):color{#c00}Rightarrow usim w$
Such relations are called (equivalence) kernels. One calls $, sim,$ the $,(approx),$ kernel of $rm,f.,$ The equivalence classes $,f_c = f^{-1}(c),$ are called the fibers or preimages, or level sets / curves of $f.$
Yours is the special case when $,approx,$ is the equivalence relation of equality.
You can find many other examples of equivalence kernels in prior answers.
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
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add a comment |
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If you rearrange the defining equation as $a-c=2b-2d$, or $frac{b-d}{a-c}=frac12$, you can find a geometric interpretation. We have $(a,b)sim (c,d)$ if and only if $(a,b)$ and $(c,d)$ lie on the same line with a slope of $frac12$. Geometrically, it is clear that this is an equivalence: any point is on the same line as itself; if point $P$ is on the same line as point $Q$, then point $Q$ is on the same line as $P$; if $P$ and $Q$ are on the same line, and so are $Q$ and $R$, then so are $P$ and $R$.
The equivalence classes under this relation are all lines with $frac12$ slope.
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
$endgroup$
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
1
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
add a comment |
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3 Answers
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3 Answers
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Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
$endgroup$
add a comment |
$begingroup$
Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
$endgroup$
Yes, your proof is perfectly correct. We can simplify the proof a little bit, especially the transitivity part, if we observe that $a+2d=c+2b$ is equivalent to $a-2b=c-2d$, as this way each side of the condition uses entries from the same pair.
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
edited Aug 11 '17 at 3:26
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
answered Aug 11 '17 at 3:20
zipirovichzipirovich
11.3k11731
11.3k11731
add a comment |
add a comment |
$begingroup$
That's a correct direct approach. More generally note that $,(a,b)sim (c,d)iff f(a,b) = f(c,d),$ for $,f(u,v) = u-2v,$ so we can apply the following very general kernel criterion.
It is straightforward to prove relations of form $rm, xsim y {overset{ def}{color{#c00}iff}} f(x) = f(y), $ are equivalence relations.
More generally, suppose $rm usim v smash[t]{overset{ def}{color{#c00}iff}}, f(u) approx f(v) $ for a function $rm,f,$ and equivalence relation $,approx., $ Then the equivalence relation $rmcolor{#0a0}{properties (E)},$ of $,approx,$ transport (pullback) to $,sim,$ along $rm,f$ as follows
reflexive $rmquad color{#0a0}{overset{(E)}Rightarrow}, f(v) approx f(v):color{#c00}Rightarrow:vsim v$
symmetric $rm, usim v:color{#c00}Rightarrow f(u) approx f(v):color{#0a0}{overset{(E)}Rightarrow}:f(v)approx f(u):color{#c00}Rightarrow:vsim u$
transitive $rm usim v,, vsim w:color{#c00}Rightarrow: f(u)approx f(v),,f(v)approx f(w):color{#0a0}{overset{(E)}Rightarrow}:f(u)approx f(w):color{#c00}Rightarrow usim w$
Such relations are called (equivalence) kernels. One calls $, sim,$ the $,(approx),$ kernel of $rm,f.,$ The equivalence classes $,f_c = f^{-1}(c),$ are called the fibers or preimages, or level sets / curves of $f.$
Yours is the special case when $,approx,$ is the equivalence relation of equality.
You can find many other examples of equivalence kernels in prior answers.
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
$begingroup$
That's a correct direct approach. More generally note that $,(a,b)sim (c,d)iff f(a,b) = f(c,d),$ for $,f(u,v) = u-2v,$ so we can apply the following very general kernel criterion.
It is straightforward to prove relations of form $rm, xsim y {overset{ def}{color{#c00}iff}} f(x) = f(y), $ are equivalence relations.
More generally, suppose $rm usim v smash[t]{overset{ def}{color{#c00}iff}}, f(u) approx f(v) $ for a function $rm,f,$ and equivalence relation $,approx., $ Then the equivalence relation $rmcolor{#0a0}{properties (E)},$ of $,approx,$ transport (pullback) to $,sim,$ along $rm,f$ as follows
reflexive $rmquad color{#0a0}{overset{(E)}Rightarrow}, f(v) approx f(v):color{#c00}Rightarrow:vsim v$
symmetric $rm, usim v:color{#c00}Rightarrow f(u) approx f(v):color{#0a0}{overset{(E)}Rightarrow}:f(v)approx f(u):color{#c00}Rightarrow:vsim u$
transitive $rm usim v,, vsim w:color{#c00}Rightarrow: f(u)approx f(v),,f(v)approx f(w):color{#0a0}{overset{(E)}Rightarrow}:f(u)approx f(w):color{#c00}Rightarrow usim w$
Such relations are called (equivalence) kernels. One calls $, sim,$ the $,(approx),$ kernel of $rm,f.,$ The equivalence classes $,f_c = f^{-1}(c),$ are called the fibers or preimages, or level sets / curves of $f.$
Yours is the special case when $,approx,$ is the equivalence relation of equality.
You can find many other examples of equivalence kernels in prior answers.
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
add a comment |
$begingroup$
That's a correct direct approach. More generally note that $,(a,b)sim (c,d)iff f(a,b) = f(c,d),$ for $,f(u,v) = u-2v,$ so we can apply the following very general kernel criterion.
It is straightforward to prove relations of form $rm, xsim y {overset{ def}{color{#c00}iff}} f(x) = f(y), $ are equivalence relations.
More generally, suppose $rm usim v smash[t]{overset{ def}{color{#c00}iff}}, f(u) approx f(v) $ for a function $rm,f,$ and equivalence relation $,approx., $ Then the equivalence relation $rmcolor{#0a0}{properties (E)},$ of $,approx,$ transport (pullback) to $,sim,$ along $rm,f$ as follows
reflexive $rmquad color{#0a0}{overset{(E)}Rightarrow}, f(v) approx f(v):color{#c00}Rightarrow:vsim v$
symmetric $rm, usim v:color{#c00}Rightarrow f(u) approx f(v):color{#0a0}{overset{(E)}Rightarrow}:f(v)approx f(u):color{#c00}Rightarrow:vsim u$
transitive $rm usim v,, vsim w:color{#c00}Rightarrow: f(u)approx f(v),,f(v)approx f(w):color{#0a0}{overset{(E)}Rightarrow}:f(u)approx f(w):color{#c00}Rightarrow usim w$
Such relations are called (equivalence) kernels. One calls $, sim,$ the $,(approx),$ kernel of $rm,f.,$ The equivalence classes $,f_c = f^{-1}(c),$ are called the fibers or preimages, or level sets / curves of $f.$
Yours is the special case when $,approx,$ is the equivalence relation of equality.
You can find many other examples of equivalence kernels in prior answers.
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
$endgroup$
That's a correct direct approach. More generally note that $,(a,b)sim (c,d)iff f(a,b) = f(c,d),$ for $,f(u,v) = u-2v,$ so we can apply the following very general kernel criterion.
It is straightforward to prove relations of form $rm, xsim y {overset{ def}{color{#c00}iff}} f(x) = f(y), $ are equivalence relations.
More generally, suppose $rm usim v smash[t]{overset{ def}{color{#c00}iff}}, f(u) approx f(v) $ for a function $rm,f,$ and equivalence relation $,approx., $ Then the equivalence relation $rmcolor{#0a0}{properties (E)},$ of $,approx,$ transport (pullback) to $,sim,$ along $rm,f$ as follows
reflexive $rmquad color{#0a0}{overset{(E)}Rightarrow}, f(v) approx f(v):color{#c00}Rightarrow:vsim v$
symmetric $rm, usim v:color{#c00}Rightarrow f(u) approx f(v):color{#0a0}{overset{(E)}Rightarrow}:f(v)approx f(u):color{#c00}Rightarrow:vsim u$
transitive $rm usim v,, vsim w:color{#c00}Rightarrow: f(u)approx f(v),,f(v)approx f(w):color{#0a0}{overset{(E)}Rightarrow}:f(u)approx f(w):color{#c00}Rightarrow usim w$
Such relations are called (equivalence) kernels. One calls $, sim,$ the $,(approx),$ kernel of $rm,f.,$ The equivalence classes $,f_c = f^{-1}(c),$ are called the fibers or preimages, or level sets / curves of $f.$
Yours is the special case when $,approx,$ is the equivalence relation of equality.
You can find many other examples of equivalence kernels in prior answers.
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
edited Jan 16 at 16:37
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
answered Aug 11 '17 at 3:21
Bill DubuqueBill Dubuque
211k29193646
211k29193646
add a comment |
add a comment |
$begingroup$
If you rearrange the defining equation as $a-c=2b-2d$, or $frac{b-d}{a-c}=frac12$, you can find a geometric interpretation. We have $(a,b)sim (c,d)$ if and only if $(a,b)$ and $(c,d)$ lie on the same line with a slope of $frac12$. Geometrically, it is clear that this is an equivalence: any point is on the same line as itself; if point $P$ is on the same line as point $Q$, then point $Q$ is on the same line as $P$; if $P$ and $Q$ are on the same line, and so are $Q$ and $R$, then so are $P$ and $R$.
The equivalence classes under this relation are all lines with $frac12$ slope.
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
$endgroup$
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
1
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
add a comment |
$begingroup$
If you rearrange the defining equation as $a-c=2b-2d$, or $frac{b-d}{a-c}=frac12$, you can find a geometric interpretation. We have $(a,b)sim (c,d)$ if and only if $(a,b)$ and $(c,d)$ lie on the same line with a slope of $frac12$. Geometrically, it is clear that this is an equivalence: any point is on the same line as itself; if point $P$ is on the same line as point $Q$, then point $Q$ is on the same line as $P$; if $P$ and $Q$ are on the same line, and so are $Q$ and $R$, then so are $P$ and $R$.
The equivalence classes under this relation are all lines with $frac12$ slope.
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
$endgroup$
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
1
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
add a comment |
$begingroup$
If you rearrange the defining equation as $a-c=2b-2d$, or $frac{b-d}{a-c}=frac12$, you can find a geometric interpretation. We have $(a,b)sim (c,d)$ if and only if $(a,b)$ and $(c,d)$ lie on the same line with a slope of $frac12$. Geometrically, it is clear that this is an equivalence: any point is on the same line as itself; if point $P$ is on the same line as point $Q$, then point $Q$ is on the same line as $P$; if $P$ and $Q$ are on the same line, and so are $Q$ and $R$, then so are $P$ and $R$.
The equivalence classes under this relation are all lines with $frac12$ slope.
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
$endgroup$
If you rearrange the defining equation as $a-c=2b-2d$, or $frac{b-d}{a-c}=frac12$, you can find a geometric interpretation. We have $(a,b)sim (c,d)$ if and only if $(a,b)$ and $(c,d)$ lie on the same line with a slope of $frac12$. Geometrically, it is clear that this is an equivalence: any point is on the same line as itself; if point $P$ is on the same line as point $Q$, then point $Q$ is on the same line as $P$; if $P$ and $Q$ are on the same line, and so are $Q$ and $R$, then so are $P$ and $R$.
The equivalence classes under this relation are all lines with $frac12$ slope.
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
answered Aug 11 '17 at 4:00
G Tony JacobsG Tony Jacobs
25.8k43685
25.8k43685
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
1
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
add a comment |
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
1
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
This is a special case of the equivalence kernel viewpoint mentioned in my answer. Here the equivalence kernel of $,f(x,y) = x-2y,$ partitions the plane into fiber (level set) equivalence classes having the form $,f_c = f^{-1}(c) = $ all points $,(x,y),$ on the line $,x-2y = c. $
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:31
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
$begingroup$
Yeah, it was reading your answer that inspired me to post this one. Thanks for the lesson on equivalence kernels! :) I just posted this separately for the geometric viewpoint, which might be accessible with or without understanding equivalence kernels in general.
$endgroup$
– G Tony Jacobs
Aug 11 '17 at 16:40
1
1
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
$begingroup$
Agreed, it is always instructive to emphasize familiar special cases of general results. I added some links on fibers to my answer.
$endgroup$
– Bill Dubuque
Aug 11 '17 at 16:43
add a comment |
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