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Dibyashanu Pati 2025-03-02 13:50:52 +05:30
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@ -168,7 +168,7 @@ Then
Since $S: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ is both injective and surjective, it is a bijection, we are done.
* An Interesting Applications
* An Interesting Application
*The set of finite subsets of $\mathbb{N}$ is countable.*
Let $A_i$ be set of all subsets of $\mathbb{N}$ containing $i$ elements, that is with cardinality $i$.