in Set Theory & Algebra retagged by
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As a refresher, if $R$ is an equivalence relation over a set $A$ and $x \in A$, then the equivalence class of $\boldsymbol{x}$ in $\boldsymbol{R}$, denoted $[x]_R,$ is the set
$$
[x]_R=\{y \in A \mid x R y\}
$$
Let's now introduce some new notation. If $R$ is an equivalence relation over a set $A,$ the index of $R,$ denoted $\boldsymbol{I}(\boldsymbol{R}),$ is the number of equivalence classes in $R$. Additionally, the width of $R,$ denoted $\boldsymbol{W}(\boldsymbol{R}),$ is the number of elements in the largest equivalence class in $R$.

If $R$ is an equivalence relation over a set $A$ and $|A|=n^2+1$ for some positive natural number $n$, then which of the following must be true?

  1. $\mathrm{I}(\mathrm{R}) \geq n+1$ and $\mathrm{W}(\mathrm{R}) \geq n+1$.
  2. $\mathrm{I}(\mathrm{R}) \leq n$ and $\mathrm{W}(\mathrm{R}) \leq n$
  3. $\mathrm{I}(\mathrm{R}) \geq n+1$ or $\mathrm{W}(\mathrm{R}) \geq n+1$ (or both).
  4. $\mathrm{I}(\mathrm{R})=n / 2$ and $\mathrm{W}(\mathrm{R})=n / 2$
in Set Theory & Algebra retagged by
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1 Answer

2 votes
2 votes

Detailed Video Solution: https://www.youtube.com/watch?v=Rfc5CavStLM&t=3567s  

If the largest equivalence class has $\leq n$ elements, then how many equivalence classes will we need?? Think.

If the number of equivalence classes is $\leq n,$ then can all equivalence classes contain $\leq n$ elements ?? Think.

Very Easy $\&$ Interesting Question... But Think.

Solution: If index of $R$ is $I$ and width of $R$ is $W,$ then $A \leq IW.$ So, if $|A| = n^2 + 1$ then at least one of $I,W$ must be $\geq n+1.$

Detailed Video Solution: https://www.youtube.com/watch?v=Rfc5CavStLM&t=3567s 

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