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Number of reflexive relations defined on a set A with n elements

I understand that for n elements of a set we have 2^(n^2-n) ways of checking whether a function is reflexive or not. But shouldn't the n confirmed reflexive relations be added to give a total of 2^(n^2-n)+n relations

Why is it that we dont include the n confirmed reflexive cases while counting

user755177
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  • Here's a tutorial and reference for typesetting math on this site. – joriki Mar 30 '20 at 08:32
  • I'm afraid you're going to have to spell that out a bit more. Where in the linked question or the answer to it do you see "$n$ confirmed reflexive cases"? – joriki Mar 30 '20 at 08:33

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You seem to be mixing up a relation, which is a subset $R\subseteq A\times A$, with their elements $(x,y)\in R$. A reflexive relation $R$ has as elements at least all pairs $(x,x)$ for $x\in A$. If $|A|=n$ this means that the $n$ elements $(x,x)$ will be in any reflexive relation. This leaves you with $n^2-n$ other possible elements of $A\times A$ that might or might not be elements of $R$. Hence the number of possible reflexive relations $R$ is $2^{n^2-n}$.

For example when $A=\{1,2,3\}$, every reflexive relation contains the elements $(1,1)$, $(2,2)$ and $(3,3)$, so $$ R = \{(1,1), (2,2), (3,3), \dots \}. $$ Now for the remaining $3^2-3 = 6$ elements $$ (1,2), (1,3), (2,1), (2,3), (3,1), (3,2) $$ they can either be in $R$ or not be in $R$, leaving us with $6$ yes/no choices. There are $2^6$ ways to make these $6$ choices.

Christoph
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