If two sets are equivalent is it ok to write they are equal

Question

I am trying to understand the proof of the following theorem:

In the very first line of the proof, it is written that

"It suffices to assume that $A_i=\mathbb{N}$"

But I do not understand why. How did the author assume that $A_i=\mathbb{N}$?

Answer 1

No, they are not equal but for all the purposes of the statement and the proof they might as well be.

We only care about the countability of the sets. As there is a bijection between $\mathbb N$ and $A_i$ anything we can say about some natural number $k$ we can say about the $k$th element of $A_i$.

To do the proof without assuming without loss of generality that $A_i = \mathbb N$ we instead declare that $\phi_i: A_i\to \mathbb N$ be a set of bijections between $A_i$ and $\mathbb N$.

Then our proof is:

Pick $n$ distinct primes $p_1,... ,p_n$ and define $f:A_1\times... A_n\to \mathbb N$ by $f(\alpha_1,\alpha_2,...., \alpha_n) = p_1^{\phi_1(\alpha_1)}p_2^{\phi_2(\alpha_2)}....p_n^{\phi_n(\alpha_n)}$. By the fundamental theorem of arithmetic that $f$ is one-to-one hence $A_1\times... A_n$ is countable.

...OR ALTERNATIVELY....

This proof proved that $\mathbb N^n$ is countable.

Let $\phi:A_1\times... A_n\to \mathbb N^n$ by $\phi(\alpha_1, \alpha_2,....,\alpha_n) = (\phi_1(\alpha_1), \phi_2(\alpha_2),...., \phi_n(\alpha_n))$. It's easy to demonstrate that $\phi$ is a bijection. So $A_1\times... A_n$ is countable.

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FWIW. A proof I would prefer is:

As $\phi:(A_1\times...\times A_{n-1})\times A_n \to A_1\times...\times A_{n-1}\times A_n$ by $\phi((a_1,....,a_{n-1}),a_n) = (a_1,...,a_{n-1}, a_n)$ is clearly a bijection.

If we can prove for any countable $A$ and $B$ that $A\times B$ is countable then $A_n \to A_1\times...\times A_{n-1}\times A_n$ is countable by induction.

Proving $A\times B$ is countable by the "diagonal" bijection is a standard graphic "proof" of any introductory text.

Answer 2

This depends on whether what you want to do is invariant under the given equivalence. In this specific case $A \cong \mathbb{N}$, $B \cong \mathbb{N}$ implies $A \times B \cong \mathbb{N} \times \mathbb{N}$. Hence it is sufficient to consider $\mathbb{N} \times \mathbb{N}$.

Answer 3

The two sets may have different structure that is used in the context. For example, the integers and the rationals have the same cardinality, yet one hardly thinks of these two sets as identical.