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So this is the question I've been stuck on: Prove the relation in $\Bbb R$ is a function . $(x,y)∈τ ⇔x^2 ≤y$;

Im used to doing this kind of questions but not with an inequality, how do I go around this?

Nick
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MooeSha123
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    How do you define function and how is a relation a function in your sense? Because how I see it this is not a function at all as for every $x$ there are many $y$ in relation to $x$... – Dirk Oct 19 '16 at 11:26
  • In that case I may have to prove that it is not a function, which I have no idea how to do with the inequality. – MooeSha123 Oct 19 '16 at 11:30
  • Well, first step would to look up and write down the definition of a function and what a relation has to fulfill to be a function. Once we have that we can start to ask if this relation obeys all the rules for a function or not. – Dirk Oct 19 '16 at 11:34
  • This is a duplicate of your previous question. As people pointed out for that one, a function, if given a value, needs to return a value. That's not the case here...what would $f(3)$ be? – lulu Oct 19 '16 at 11:48

2 Answers2

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In general you could prove a relation "is" a function, by showing that every x has one and only one tuple $(x,y) \in \tau$. That however is not the case, since e.g. $(1,2) \in \tau$ and $(1,3) \in \tau$.

The relation however could be a function in a adequate set e.g. $\{1\} \rightarrow \{1\}$

don-joe
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  • Yh thats what I was thinking, so is it okay to just give an example to show that for some some x there are multiple y's in this inequality. – MooeSha123 Oct 19 '16 at 11:41
  • whenever "forall"-condition is supposedly not fulfilled a counterexample is the way to go. so: yes. – don-joe Oct 19 '16 at 12:25
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If you have a relation $R':=\{A,B,G\}$ where $G\in A\times B$. Then $R'$ is said to be a function if the followings hold:
(a)$\forall x\in A, \exists y\in B$ such that $(x,y)\in G$, and
(b)$\forall x\in A, \forall y,z\in B$, if $(x,y),(x,z)\in G$ then $y=z$.

So in your case, we check (a) and (b) one by one:

Let the relation $R:=\{\Bbb R,\Bbb R, F\}$, where $F:=\{(x,y):x,y\in \Bbb R \text{ and }x^2\le y\}$.

Let $x\in \Bbb R$, then $(x,x^2)\in F$. (a) holds.
Let $x,y,z\in \Bbb R$ such that $(x,y),(x,z)\in F$, then $y$ may not be $z$ , one simple example would be $(x,x^2),(x,(x+1)^2)\in F$ with $x^2\neq (x+1)^2$. (b) does not hold.

So $R$ is not a function.

Nick
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