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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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


















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







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 5:06









anomaly

17.5k42664




17.5k42664










asked Dec 10 '18 at 5:01









CColaCCola

346




346












  • $begingroup$
    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










  • $begingroup$
    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












1 Answer
1






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 $$x_1 = x_2 implies y_1=y_2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






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 $$x_1 = x_2 implies y_1=y_2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















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 $$x_1 = x_2 implies y_1=y_2$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














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 $$x_1 = x_2 implies y_1=y_2$$






share|cite|improve this answer











$endgroup$



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 $$x_1 = x_2 implies y_1=y_2$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 11:54

























answered Dec 10 '18 at 5:08









Mohammad Riazi-KermaniMohammad Riazi-Kermani

41.5k42061




41.5k42061












  • $begingroup$
    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










  • $begingroup$
    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


















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