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We define the relationship between the two sets $S_1≡S_2$ if and only if $|S_1|=|S_2|$.

How to show that $≡$ is an equivalence relation ?

sorry I'm from Iran and Basic my English is poor.

Mauro ALLEGRANZA
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shiva
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  • Technically, it is not a relation at all, because a relation is defined on a set, and the collection of all sets is not a set. I'm sure there is some name for an equivalence like this, but it isn't coming to me. – Thomas Andrews Jun 19 '15 at 20:30
  • @ThomasAndrews equinumerosity ? when two sets have same cardinality, which is my understanding of the question. – lmsteffan Jun 19 '15 at 20:33
  • You have to use the definition of $|A|=|B|$ in Cardinality of a set and show that it satisfies the three properties of an Equivalence relation : "A binary relation is said to be an equivalence relation if and only if it is reflexive, symmetric and transitive." – Mauro ALLEGRANZA Jun 19 '15 at 20:33
  • @lmsteffan I didn't mean this relation, but the general idea of an equivalence that is not a relation. – Thomas Andrews Jun 19 '15 at 20:35
  • I believe that the issue that @thomas-andrews is referring to is the fact that an equivalence relation is defined on a set. If you have some restricted set of sets, then you can work in the usual way, but the class of all sets is not a set. – copper.hat Jun 19 '15 at 21:12

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we say that $|s_1|=|s_2|$ if there is a funtion $f:S_1\rightarrow S_2$ such that $f$ is a bijection, so Is clear that $S\cong S$ putting $f=I_s$ where $I_s(x)=x, \forall x \in S$, now if $S_1\cong S_2$ then $S_2\cong S_1$ ussing like bijecction $f^{-1}:S_2\rightarrow S_1$, and finally if $S_1\cong S_2$ and $S_2\cong S_3$ there are bijeecion $f,g$ and doing the composition $g\circ f $ we get a bijecction between $S_1$ and $S_3$ so $S_1\cong S_3$, formally this is not a relation because the realction are definided over sets and this not, but I think tah this is that you want. the best :)

sti9111
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  • Thank you very much. it is Question 4 The first chapter of the book of languages and automata theory is Peter Linz. Your answer to this question, right? – shiva Jun 20 '15 at 12:42
  • I do not understand the questions related to the Mathematical induction of this book!! Can you help me?There's a good reference. – shiva Jun 20 '15 at 12:49
  • do the question . – sti9111 Jun 20 '15 at 17:46