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I. $\{(1,1),(2,2),(3,3),(4,4), (5, 5), (1,3), (3,4)\}$ is reflexive

II. $\{(1,1), (2,2), (3,3), (4,4), (5,5)\}$ is reflexive, symmetrical, transitive

Why the second set is symmetrical, when we don't have, for example, $(1, 2)$ and $(2, 1)$?

Why the second set is transitive, when we don't have, for example, $(1,3)$ and $(3, 4)$ and $(1, 4)$?

Georgi Peev
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    Just because it hasn't got every possible symmetry, doesn't mean it is not symmetric. – Nij Oct 23 '21 at 01:43
  • @Nij It doesn't have at least one symmetric what about all ... Look at the second one it doesn't have for example (1, 2) & (2, 1) but it is symmetrical, the first one also doesn't have the same pattern. – Georgi Peev Oct 23 '21 at 01:47
  • @Nij If we count (1,1) as symmetrical then why the first set is not symmetrical? – Georgi Peev Oct 23 '21 at 01:47
  • I don't think you understand what symmetric means, then. – Nij Oct 23 '21 at 01:48

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Symmetry means that if the set contains $(a,b)$, then it must contain $(b,a)$ as well. But that does not mean that it contains both $(a,b)$ and $(b,a)$ ... it could also contain neither.

Put differently, the only way for a set to be not symmetric, is if it contains $(a,b)$ for some $a$ and $b$, but not $(b,a)$. This is not the case for the second set, so the second set is symmetric. The first set is not symmetric, since it contains $(1,3)$, but not $(3,1)$

Likewise, transitivity means that if the set contains $(a,b)$ and $(b,c)$, then it contains $(a,c)$ as well. For a set not to be transitive it has to contain $(a,b)$ and $(b,c)$, but not $(a,c)$ for some $a$, $b$, and $c$. Again, this is not the case for the second set, so the second set is transitive. But the first set is not transitive: it contains $(1,3)$ and $(3,4)$, but not $(1,4)$.

Bram28
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