Cardinality of infinite subset of infinite set

AI Thread Summary
Cardinality refers to the size of sets, and if A is a subset of B, then the cardinality of A is less than or equal to that of B. The discussion clarifies that while it is true that |A| ≤ |B| for subsets, it does not imply that |A| is strictly less than |B|. Specifically, the cardinalities of the integers (ℤ) and the rationals (ℚ) are equal, despite the intuitive assumption that ℤ is smaller. A bijection can be established between ℕ × ℕ and ℕ, which helps illustrate the equality of cardinalities. Understanding these relationships is crucial in set theory and the study of infinite sets.
Bipolarity
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Am a bit confused about the meaning of cardinality. If ## A \subseteq B ##, then is it necessarily the case that ## |A| \leq |B| ##?

I am thinking that since ## A \subseteq B ##, an injection from A to B exists, hence its cardinality cannot be greater than that of B?

But this cannot be correct, since ##\mathbb{Z}## and ##\mathbb{Q}## have the same cardinality?

Where am I wrong?

Thanks!

BiP
 
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You are right. If ##A\subseteq B##, then ##|A|\leq |B|##.

In particular, ##|\mathbb{Z}|\leq |\mathbb{Q}|## is true. Don't confuse this with ##|\mathbb{Z}|< |\mathbb{Q}|##, which is false.
 
R136a1 said:
You are right. If ##A\subseteq B##, then ##|A|\leq |B|##.

In particular, ##|\mathbb{Z}|\leq |\mathbb{Q}|## is true. Don't confuse this with ##|\mathbb{Z}|< |\mathbb{Q}|##, which is false.

I see. So which of the following is true?
##|\mathbb{Z}|< |\mathbb{Q}|##
##|\mathbb{Z}|= |\mathbb{Q}|##

Thanks!

BiP
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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