1

I am not quite sure how to show the following:

Let $\mathbb{Z}^{\oplus \mathbb{N}}$ be set of functions $f:\mathbb N \to \mathbb Z$ such that $f(i) \ne 0$ for finitely many $i\in \mathbb N$. Let $\mathbb Z^{\mathbb N}$ be set of functions from $\mathbb N \to\mathbb Z$. Show that $\mathbb{Z}^{\mathbb{N}}$ is not isomorphic to $\mathbb{Z}^{\oplus \mathbb{N}}$.

So far, I know that $\mathbb{Z}^{\oplus \mathbb{N}}$ is generated by $f_i:\mathbb N \to \mathbb Z$ for $i \in \mathbb N$ where $f_i(x)=1$ if $x=i$ and $0$ otherwise. In other words, every $f\in \mathbb{Z}^{\oplus \mathbb{N}}$ can be represented as a finite sum $\sum m_i f_i$ where $m_i\in \mathbb Z$. So the question above is equivalent to asking that whether $\mathbb Z^{\mathbb N}$ is generated by a countable set of functions $f:\mathbb N \to \mathbb Z$.

Any help is appreciated. Thanks!

user26857
  • 52,094
Tengu
  • 4,072

1 Answers1

5

Isomorphisms are defined to be bijections. So if two algebraic structures are isomorphic their corresponding carrier sets have the same cardinality.

Starting with

1) $\text{#}( \mathbb{Z}^{\oplus \mathbb{N}}) = \aleph _{0}$.
2) $\aleph _{0} \lt \text{#}(2^{\Bbb N}) = \mathfrak {c}$

use celtschk's hint

Construct an injection from $ 2^{\Bbb N}$ into $\mathbb{Z}^{\mathbb{N}}$ to show that $\mathfrak {c} \le \text{#}(\mathbb{Z}^{\mathbb{N}})$.

Note that by using a forgetful functor, the problem of analyzing algebraic morphisms melts away.

The OP can show that $\mathbb{Z}^{\oplus \mathbb{N}}$ is countable by using the following technical machinery
(see Wikipedia: Countable set):

Proposition: If $A_n$ is a countable set for each $n \in \Bbb N$ then the union of all $A_n$ is also countable.

and defining for each $n \in \Bbb N$ the set

$$ F_n = \bigr\{\sum_{k=0}^n m_k f_k \, | \, -n \le m_k \le n \bigr\}$$

CopyPasteIt
  • 11,366
  • With my limited knowledge about set theory, I am unable to show $\mathbb{Z}^{\oplus \mathbb N}$ is countable. Could you please give me some hint for this part? Thank you! (I did notice a link given by @DietrichBurde stating that $A$ and $\langle A\rangle$ have the same cardinality but I don't really understand the arguments given in the link). – Tengu Jul 09 '19 at 00:52
  • 1
    @Tengu: Another way to see that $\mathbb Z^{\oplus\mathbb N}$ is countable is to use the uniqueness of prime factorization: When $p_n$ is the sequence of primes, then the function $(a_n)\mapsto \prod_n p_n^{a_n}$ is an injection from $\mathbb Z^{\oplus\mathbb N}$ into $\mathbb N$. – celtschk Jul 13 '19 at 10:51
  • 1
    @Tengu: Actually I just notice that as written, my comment is wrong: It would be true for $\mathbb N^{\oplus\mathbb N}$. So you first have to map $\mathbb Z$ to $\mathbb N$, and use that mapping in the exponent. But fortunately that's not hard. Alternatively you can observe that the function as given is an injection into $\mathbb Q$, which also is countable. – celtschk Jul 13 '19 at 11:14