To prove there is no bijection between a natural number set $$x$$ and its successor $$x^+=x\cup\{x\}$$, one must show that the cardinality of $$x$$ is less than that of $$x^+$$. Assuming a bijection exists leads to a contradiction, as it implies $$x$$ is infinite, which contradicts the properties of natural numbers. The argument relies on the fact that $$x$$ cannot be an element of itself, reinforcing that there is no one-to-one correspondence. Thus, it can be concluded that $$\mathrm{card}\,x<\mathrm{card}\,x^+$$ holds true. This establishes the foundational understanding of natural number sets and their successors in set theory.
