Given that , A and B are two set with same cardinality .i,e, ∣A∣=∣B∣
Claim: ∣A∣≤∣B∣ and ∣B∣≤∣A∣
Since, cardinality of A and B are equal .Therefore there exist a bijective function f:A→B
Since f is one-one and f(A)⊆B
∴∣A∣≤∣B∣
Again , f−1:B→A is a bijective map .
Therefore f−1 is one-one and f−1(B)⊆A
Hence ,
∣B∣≤∣A∣
Comments