MHB Prove No Bijection between x and x^+

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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.
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Let $$x$$ be a natural number (set). How to prove that there is no bijection between $$x$$ and $$x^+$$, where $$x^+=x\cup\{x\}$$? Then I can show that $$\mathrm{card}\,x<\mathrm{card}\,x^+.$$ I know that $$x\notin x.$$
 
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Andrei said:
$$\mathrm{card}\,x<\mathrm{card}\,x^+.$$
I took this problem from Zorich "Mathematical analysis". Zorich spoke just twice about infinity before this problem: 1. Dedekind's definition, 2. Axiom of infinity. Suppose there is a bijection from $$x$$ to $$x^+.$$ Since $$x\subset x^+$$, then $$x$$ is infinite by 1, which is false, but I don't know how to prove it.
 

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