In various other words: is it a requisite the there exist at the very least one notified pair the the type \$(x,y)\$ for every \$xin X\$. Where \$xin X\$, \$yin Y\$ and also the relation is of the form \$XrY\$?

Yes. The meaning of a function is a relation which has second property: if both \$(x,y)\$ and \$(x,z)\$ room in the relation climate \$y=z\$.

You are watching: Every relation is a function.

More to the point, though, the notation \$fcolon X o Y\$ method three things:

\$f\$ is a function, which is a relation with this one-of-a-kind property.\$operatornamedom(f)=X\$, therefore every suggest in \$x\$ appears as the left name: coordinates of part ordered pair in \$f\$.\$operatornamerng(f)subseteq Y\$, therefore if \$(x,y)in f\$ then \$yin Y\$.

There room contexts wherein we eliminate the second requirement. For example in forcing we frequently want to talk around partial features from a huge domain, so us omit the 2nd requirement. Yet this is normally mentioned clearly in the text.

As Halmos says, view a relationship (in specific a function) in terms of something the is, no something that does.

A relationship is a subset of \$X imes Y\$. That is all. You choose pairs \$(x,y)\$ with \$xin X,yin Y\$. If \$(x,y)in R\$; then \$x\$ stands in relation with \$y\$.

A role is a special type of relation: all aspects of \$X\$ need to be concerned at least one element of \$Y\$. Moreover, this facet \$xin X\$ is regarded is unique, that is, \$x\$ cannot be associated to more than one aspect of \$Y\$.

So, for example, a function \$f:1,3,6,5\toa,b,c\$ is \$\$f=(1,a),(3,b),(6,b),(5,c)\$\$

while a relation can be \$\$R=(1,a),(1,b),(3,b),(3,c)\$\$

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