What properties does $R$ have to satisfy for $R=operatorname{graph}(f)$ for some function $f:Ato B$?
0
$begingroup$
Given a relation $R$ between $A$ and $B$, what properties does $R$ have to satisfy for $R=operatorname{graph}(f)$ for some function $f:Ato B$?
relations
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
asked Dec 10 '18 at 5:01
CColaCCola
376
$endgroup$
add a comment |
0
$begingroup$
Given a relation $R$ between $A$ and $B$, what properties does $R$ have to satisfy for $R=operatorname{graph}(f)$ for some function $f:Ato B$?
relations
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
asked Dec 10 '18 at 5:01
CColaCCola
376
$endgroup$
$begingroup$
To every elements in $a in A$, there is a unique element in $b in B$ such that $aRb$?
$endgroup$
– nafhgood
Dec 10 '18 at 5:04
$begingroup$
What does "graph(f)" mean?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:11
$begingroup$
@Acccumulation The graph of a function $f:A to B$ is the set ${(x, f(x)) : x in A }$.
$endgroup$
– Patrick Stevens
Dec 10 '18 at 5:19
add a comment |
0
0
0
$begingroup$
Given a relation $R$ between $A$ and $B$, what properties does $R$ have to satisfy for $R=operatorname{graph}(f)$ for some function $f:Ato B$?
relations
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
asked Dec 10 '18 at 5:01
CColaCCola
376
$endgroup$
Given a relation $R$ between $A$ and $B$, what properties does $R$ have to satisfy for $R=operatorname{graph}(f)$ for some function $f:Ato B$?
relations
relations
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
asked Dec 10 '18 at 5:01
CColaCCola
376
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
asked Dec 10 '18 at 5:01
CColaCCola
376
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
edited Dec 10 '18 at 5:06
anomaly
17.7k42666
17.7k42666
asked Dec 10 '18 at 5:01
CColaCCola
376
asked Dec 10 '18 at 5:01
CColaCCola
376
asked Dec 10 '18 at 5:01
CColaCCola
376
376
$begingroup$
To every elements in $a in A$, there is a unique element in $b in B$ such that $aRb$?
$endgroup$
– nafhgood
Dec 10 '18 at 5:04
$begingroup$
What does "graph(f)" mean?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:11
$begingroup$
@Acccumulation The graph of a function $f:A to B$ is the set ${(x, f(x)) : x in A }$.
$endgroup$
– Patrick Stevens
Dec 10 '18 at 5:19
add a comment |
$begingroup$
To every elements in $a in A$, there is a unique element in $b in B$ such that $aRb$?
$endgroup$
– nafhgood
Dec 10 '18 at 5:04
$begingroup$
What does "graph(f)" mean?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:11
$begingroup$
@Acccumulation The graph of a function $f:A to B$ is the set ${(x, f(x)) : x in A }$.
$endgroup$
– Patrick Stevens
Dec 10 '18 at 5:19
$begingroup$
To every elements in $a in A$, there is a unique element in $b in B$ such that $aRb$?
$endgroup$
– nafhgood
Dec 10 '18 at 5:04
$begingroup$
To every elements in $a in A$, there is a unique element in $b in B$ such that $aRb$?
$endgroup$
– nafhgood
Dec 10 '18 at 5:04
$begingroup$
What does "graph(f)" mean?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:11
$begingroup$
What does "graph(f)" mean?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:11
$begingroup$
@Acccumulation The graph of a function $f:A to B$ is the set ${(x, f(x)) : x in A }$.
$endgroup$
– Patrick Stevens
Dec 10 '18 at 5:19
$begingroup$
@Acccumulation The graph of a function $f:A to B$ is the set ${(x, f(x)) : x in A }$.
$endgroup$
– Patrick Stevens
Dec 10 '18 at 5:19
add a comment |
1 Answer
1
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oldest
votes
0
$begingroup$
You need two properties.
1) For each $xin A$ there is a $yin B$ such that $(x,y)in R$
2) If $(x_1,y_1)in R$ and $(x_2,y_2)in R$, then $$x_1 = x_2 implies y_1=y_2$$
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
0
$begingroup$
You need two properties.
1) For each $xin A$ there is a $yin B$ such that $(x,y)in R$
2) If $(x_1,y_1)in R$ and $(x_2,y_2)in R$, then $$x_1 = x_2 implies y_1=y_2$$
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
add a comment |
0
$begingroup$
You need two properties.
1) For each $xin A$ there is a $yin B$ such that $(x,y)in R$
2) If $(x_1,y_1)in R$ and $(x_2,y_2)in R$, then $$x_1 = x_2 implies y_1=y_2$$
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
add a comment |
0
0
0
$begingroup$
You need two properties.
1) For each $xin A$ there is a $yin B$ such that $(x,y)in R$
2) If $(x_1,y_1)in R$ and $(x_2,y_2)in R$, then $$x_1 = x_2 implies y_1=y_2$$
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
$endgroup$
You need two properties.
1) For each $xin A$ there is a $yin B$ such that $(x,y)in R$
2) If $(x_1,y_1)in R$ and $(x_2,y_2)in R$, then $$x_1 = x_2 implies y_1=y_2$$
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
edited Dec 10 '18 at 11:54
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
answered Dec 10 '18 at 5:08
Mohammad Riazi-KermaniMohammad Riazi-Kermani
41.6k42061
41.6k42061
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
add a comment |
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
I think the second point is a bit unclear. Perhaps rewrite it as "If $(x,y_1), (x,y_2) isin R$ then $y_1=y_2$"?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:25
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
$begingroup$
Thanks, I have edited my answer.
$endgroup$
– Mohammad Riazi-Kermani
Dec 10 '18 at 11:54
add a comment |
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$begingroup$
To every elements in $a in A$, there is a unique element in $b in B$ such that $aRb$?
$endgroup$
– nafhgood
Dec 10 '18 at 5:04
$begingroup$
What does "graph(f)" mean?
$endgroup$
– Acccumulation
Dec 10 '18 at 5:11
$begingroup$
@Acccumulation The graph of a function $f:A to B$ is the set ${(x, f(x)) : x in A }$.
$endgroup$
– Patrick Stevens
Dec 10 '18 at 5:19