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For basically any equivalence relation that I've encountered in my mathematical studies, the only problematic part (if at all) was proving transitivity. Is there a "real-world" (i.e. appearing in some field of mathematics) example of an equivalence relation where the reverse is the case? That is, an equivalence relation such that reflexivity and/or symmetry is not easy to see (and preferably one where transitivity is trivial on the other hand).

One such example can be found at the end of this answer, but it seems more like an illustration than something one actually encounters in one's studies.

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  • $\begingroup$ Not related to equivalence relations, but this is my favoritte example of a symmetric relation that is far from trivial to prove. $\endgroup$ – Winther Mar 10 '15 at 5:11
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That depends on what does "real-world" mean for you.

For example, in graph theory there is a non-trivial property of matrices such that the matrix has the property if and only if the associated graph is strongly connected (see here and here). (I'm using property here, not the actual name, because any property that implies strong connectivity would do, another could be an obfuscated way of saying that a directed graph is Eulerian.)

Now you can set things up like follows: suppose that $A$ is the matrix with the above property and $G$ is the associated graph. Define your relation $R \subset V\times V$ as $$v_1\ R\ v_2 \iff \text{ there is a non-zero length walk from $v_1$ to $v_2$ in }G.$$ This defines an equivalence relation where the classes of equivalence are the strongly connected components. However, it is not trivial to prove. On the other hand transitivity is obvious, because it's very easy to construct a walk (i.e. possibly non-simple path) from $v_1$ to $v_3$ given the walks from $v_1$ to $v_2$ and $v_2$ to $v_3$.

I hope this helps $\ddot\smile$

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