Is this relation reflexive, symmetric and transitive?

Question

Define a relation $R$ on the set of functions from $\mathbb{R}$ to $\mathbb{R}$ as follows:

$(f, g) \in R $ if and only if $f(x) - g(x) \geq 0$ for all $x \in \mathbb{R}$

Is this relation reflexive? symmetric? transitive? Is it an equivalence relation? Explain.

I know that for it to be an equivalance relation, it has to be reflexive, symmetric and transitive. However I am not sure what it means by "on the set of functions".

A formal solution would help:)

Cheers,

Chris

Answer 1

Reflexive:

For every function $f(x)$, $f:\mathbb R \to \mathbb R$, can we say that $f(x) - f(x) \geq 0 \forall x \in \mathbb R$?

If so then the relation is reflexive.

Symmetric:

If $f(x), g(x)$ are functions from $\mathbb R \to \mathbb R$, and IF $f(x) - g(x) \geq 0$, can we conclude that $g(x) - f(x)\geq 0 \forall x \in \mathbb R\;?$

If not, the relation is not symmetric. For a formal proof, you need only give a counterexample to prove a property doesn't hold.

Transitive:

If $f(x), g(x), h(x)$ are functions from $\mathbb R \to \mathbb R$, and IF $f(x) - g(x) \geq 0\;$ and $\;g(x)- h(x) \geq 0,\;$ does it necessarily follow that $\;f(x)- h(x) \geq 0\;?$

If so, the relation is transitive.

Answer 2

It is reflexive because the difference between a function and itself is $0$.

It is not symmetric. Take $f(x) = 1$ and $g(x) = 0$. Then $(f, g) \in R$ but $(g, f) \notin R$.

R is transitive.

If $(f,g) \in R$, then $f$ is greater than or equal to $g$.

If $(g, h) \in R$, then $g$ is greater than or equal to $h$.

The "greater than or equal to" relation is transitive, so $f$ is greater than or equal to $h$, i.e. $(f, h) \in R$.

Answer 3

For notational simplicity, we often use the notation ${}^AB$ to denote the set of all functions $f:A\to B.$

So, $R$ is a relation on ${}^{\Bbb R}\Bbb R$ given by $(f,g)\in R$ if and only if $f(x)-g(x)\ge0$ for all $x\in\Bbb R.$

In order for $R$ to be reflexive on ${}^{\Bbb R}\Bbb R,$ we must be able to say that $(f,f)\in R$ for all $f\in{}^{\Bbb R}\Bbb R,$ meaning that if $f:\Bbb R\to\Bbb R,$ then $f(x)-f(x)\ge 0$ for all $x\in\Bbb R.$ Is this true? If not, what would be a counterexample function?

In order for $R$ to be symmetric on ${}^{\Bbb R}\Bbb R,$ we must be able to say that if $(f,g)\in R$ for some $f,g\in{}^{\Bbb R}\Bbb R,$ then $(g,f)\in R.$ That is, if $f,g:\Bbb R\to\Bbb R$ are such that $f(x)-g(x)\ge0$ for all $x\in\Bbb R,$ then $g(x)-f(x)\ge 0$ for all $x\in\Bbb R.$ Is this true? If not, what would be a counterexample pair of functions?

In order for $R$ to be transitive on ${}^{\Bbb R}\Bbb R,$ we must be able to say that if $(f,g)\in R$ and $(g,h)\in R$ for some $f,g,h\in{}^{\Bbb R}\Bbb R,$ then $(f,h)\in R.$ That is, if $f,g,h:\Bbb R\to\Bbb R$ are such that $f(x)-g(x)\ge0$ and $g(x)-h(x)\ge0$ for all $x\in\Bbb R,$ then $f(x)-h(x)\ge 0$ for all $x\in\Bbb R.$ Is this true? If not, what would be a counterexample trio of functions?

For $R$ to be an equivalence relation on ${}^{\Bbb R}\Bbb R$ is the same as being reflexive, symmetric, and transitive. If it fails one or more of the above, then it is not an equivalence relation. If all of the above hold, then it is an equivalence relation.