# Proving that exists equivalence relation $r$ in set $A$ such that $|A \setminus r| = n$

I am trying to show that if $$|A| = m$$ and $$0\neq n \le m$$ then exists equivalence relation $$r$$ in set $$A$$ such that $$|A \setminus r| = n$$. Could someone help me deal with it?

• Can't you just arbitrarily glue points together? – Klaus Jan 21 at 1:15
• I don't understand, could you explain it? – René Accardo Jan 21 at 1:18
• Draw your $m$ points on a sheet of paper. Now draw lines between the points until there are exactly $n$ connected components. This defines an equivalence relation. – Klaus Jan 21 at 1:21
• Is it reflexive? – René Accardo Jan 21 at 1:33
• Added some details as an answer. – Klaus Jan 21 at 1:48

Without loss of generality let $$A = \{1, \ldots, m\}$$. Choose $$A_1 = \{1\}$$, $$A_2 = \{2\}$$, ..., $$A_{n-1} = \{n-1\}$$, $$A_n = \{n, \ldots, m\}$$. This defines a partition of $$A$$ and hence and equivalence relation.