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Given two binary relations $R$ and $S$ over sets $A$ and $B$,then $R$ is said to be contained in $S$ if $$\forall a,b: (a,b) \in R \implies (a,b) \in S$$

Moreover $R$ is considered to be smaller than $S$ if $R$ is contained in $S$,but $S$ is not contained in $R$,e.g.$$R ⊊ S$$

Wikipedia gives an example:

On the rational numbers, the relation $>$ is smaller than $≥$, and equal to the composition $> ∘ >$.

I don't understand the example,why $>$ is smaller than $≥$?

And what is the composition relation given by $> ∘ >$?

  • Can you prove that $>$ is contained in $\geq$? How about that $\geq$ is not contained in $>$? Hint: in the first part, let $a,b\in\mathbb Q$ such that $a>b$. For the second part, you need to find $a,b\in\mathbb Q$ such that $a\geq b$ but $a\not>b$. –  Jul 31 '20 at 06:03

2 Answers2

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It is easy to see that relation $\gt$ is smaller than $\geq$.

It translates to x $\gt$ y implies x $\geq$ y (and not the other way around). You can easily prove this as relations are just sets.

Composition of relations is a very interesting topic. It is a general form of composition of functions. Check out its definition here

In this particular case, you can use the fact that there exists another rational number between any two different rational numbers.

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I don't understand the example,why > is smaller than ≥?

If $a>b$ then we also have $a\geq b$, so $>$ is contained in $\geq$. Additionally, $1$ is a rational number such that $1 \geq 1$ but $1\not>1$, so $>$ is smaller than $\geq$.

And what is the composition relation given by >∘>?

By transitivity, $a > b$ and $b > c$ implies $a > c$ so it follows that $> \circ >$ is contained in $>$. On the other hand, if $a>b$ where $a,b$ are rational, then $m = \frac{a+b}{2}$ is also rational and $a > m > b$, so $>$ is contained in $> \circ >$. Therefore, $> \circ >$ equals $>$.

  • Yes, that $c$ was a typo. Any rational number between $a$ and $b$ would work for $m$, but we need to pick one for the purposes of the proof. – Brian Moehring Jul 31 '20 at 08:41