Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $R \subseteq A \times A$ is it true that $R$ is symmetrical since $xRy$ then $yRx$ ?

I have written that this is also antisymmetrical if both $x\leq y$ and $y\leq x$ if $x =y$

How does I then make $R \subseteq A \times A$ transitive?

share|cite|improve this question
In English, it's called "transitive". – Arturo Magidin Mar 16 '11 at 21:38

You are either confused, or we are running against a language barrier (both are certainly possible).

It is not true that simply by virtue of being a subset of $A\times A$, a subset $R$ will be "symmetrical."

Rather: we define $R$ to be "symmetrical" if and only if for every $x$ and $y$ in $A$, if $(x,y)\in R$ (that is, if $xRy$), then $(y,x)\in R$ (that is, $yRx$).

An example of a symmetric relation on $\mathbb{R}$ is $$R = \{ (a,b)\in\mathbb{R}\times\mathbb{R} \mid |a|=|b|\}$$ since, if $aRb$, then $|a|=|b|$, hence $|b|=|a|$, hence $bRa$.

An example of a non-symmetric relation on $\mathbb{R}$ is $$S = \{(a,b)\in\mathbb{R}\times\mathbb{R} \mid a\geq 0\}.$$ It is not symmetric, because $(1,-1)\in S$, but $(-1,1)\notin S$.

We define $R$ to be "antisymmetric" if and only if for every $x$ and $y$ in $A$, if $xRy$ and $yRx$ are both true, then $x=y$.

(You have the implication reversed).

In the example $R$ above, $R$ is not antisymmetric, because $(1,-1)\in R$, $(-1,1)\in R$, but $-1\neq 1$.

However, an example of an antisymmetric relation on $\mathbb{R}$ is: $$T = \{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a\leq b\}.$$ This is antisymmetric because if $(a,b)\in T$ and $(b,a)\in T$, that means that $a\leq b$ and $b\leq a$, so then we conclude that $a=b$.

Finally, a relation $R\subseteq A\times A$ is said to be "transitive" if and only if for every $a,b,c\in A$, if $aRb$ and $bRc$ both hold, then $aRc$ holds.

Both $R$, $S$, and $T$ above are transitive. To see an example of a relation on $\mathbb{R}$ that is not transitive, let $$U = \{(a,b)\in\mathbb{R}\times\mathbb{R}\mid a\neq b\}.$$ Then $0U1$ holds, and $1U0$ holds, but $0U0$ does not hold. So $U$ is not transitive.

share|cite|improve this answer
I think it was both language barrier, and I was confused by symmetric and antisymmetric meaning. It now makes more sense. Its the small things like this that make the rest of the path. Thank you. – user8322 Mar 16 '11 at 21:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.