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 \(\displaystyle x\) to \(\displaystyle x^+.\) Since \(\displaystyle x\subset x^+\), then \(\displaystyle x\) is infinite by 1, which is false, but I don't know how to prove it.Andrei said:\(\displaystyle \mathrm{card}\,x<\mathrm{card}\,x^+.\)
A bijection is a function between two sets that is both injective (one-to-one) and surjective (onto). This means that each element in the first set is paired with a unique element in the second set, and every element in the second set has a corresponding element in the first set.
There cannot be a bijection between x and x^+ because the sets have different cardinalities. The set x^+ includes all positive integers, while the set x includes only the number x. Since there are infinitely many positive integers, there is no way to pair each element in x with a unique element in x^+.
No, even if x is a negative number, there cannot be a bijection between x and x^+. This is because x^+ still includes all positive integers, and there is no way to pair each element in x with a unique element in x^+.
The cardinality of x is 1, as there is only one element in the set. The cardinality of x^+ is infinite, as there are infinitely many positive integers in the set. This difference in cardinality is what prevents there from being a bijection between the two sets.
No, the same principle applies even if x is a non-integer number. The set x^+ will still include all positive integers, and there will be no way to pair each element in x with a unique element in x^+ due to the infinite cardinality of x^+.