From the example of binary-symmetric-relation demonstrated in Wikipedia, how can they say the relation "$x$ and $y$ are odd numbers" is symmetric without stating any set of $x$, $y$? If such set is $\mathbb{N}$, then the relation is not symmetric because relation set does not contain even numbers. Am I thinking correctly or just missing something?
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Welcome to MSE :-) – Stefan Hamcke Nov 24 '13 at 22:43
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By the way the chat here also supports LateX, but you need some javascript bookmarks to make it work, see http://meta.math.stackexchange.com/questions/1088/should-chat-have-tex-support/3297#3297 – Stefan Hamcke Nov 24 '13 at 22:46
1 Answers
The relation is defined as $x\sim y$ if $x,y$ are both odd numbers. They do not specify the set $X$, on the other hand, this $X$ can be whatever you want. It could be the set of naturals, or the set of all real numbers, or even the set of all topologies on some set.
To check that this is indeed symmetric, let $x\sim y$. Then $x$ and $y$ are odd numbers. But then we can also say "$y$ and $x$ are odd numbers", hence $y\sim x$.
You can visualize $x\sim y\iff x,y\in A$ as a subset of $X\times X$ by thinking of the Cartesian square $(A\cap X)\times(A\cap X)$ within $X×X$. This way one sees easily that it is symmetric and transitive, and it's reflexive only if $A\supseteq X$. That is why normally we add to this relation the diagonal in $X×X$. We then get the equivalence relation $x∼y\iff x=y\vee x,y\in A$.
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