# Question about direct product.

A classic example in algebraic topology is The Shrinking Wedge of Circles $$X$$, which is the union of circles $$C_n$$ of radius $$\frac{1}{n}$$ and center $$(\frac{1}{n},0 )$$ for $$n=1,2,3,…$$.

In Hatcher’s, it says that the product of surjections $$\rho_n:\pi_1 (X) \to \pi _1(C_n)$$ gives a surjective homomorphism $$\rho:\pi_1(X) \to \prod_\infty Z$$. And since $$\prod_\infty Z$$ is uncountable, $$\pi_1(X)$$ is uncountable.

I can’t understand why $$\prod_\infty Z$$ is uncountable? Why is the direct product of countable groups uncountable?

• what is this $Z$? – Alvin Lepik Jul 30 '19 at 8:53

Suppose that $$G = \prod_{n \in \mathbb{N}} \mathbb{Z}$$ is countable. Then we can write $$G = \lbrace a_0, a_1, a_2 \dots \rbrace$$ (i.e. choose a bijection to the natural numbers), where $$a_i = (b_{i,0},b_{i,1},b_{i,2} \dots)$$ is a sequence in the integers. We can construct a sequence in the integers which is not given by $$a_n$$ for any $$n \in \mathbb{N}$$: Let $$(c_n)_n$$ be a sequence given by choosing $$c_0 \neq b_{0,0}$$, $$c_1 \neq b_{1,1}$$ etc. Thus $$c \neq a_n$$ for all $$n \in \mathbb{N}$$, a contraction.
To see this, let $$A_n$$ be a set with at least two elements for each $$n \in \mathbb{N}$$. We want to show that $$\prod_{n \in \mathbb{N}} A_n$$ is uncountable infinite.
So let $$f: \mathbb{N} \to \prod_{n \in \mathbb{N}} A_n$$ be a map. We shall show that it is not surjective. For each $$n \in \mathbb{N}$$, there exists $$a_n \in A_n$$ such that $$f(n)_n \neq a_n$$ because $$A_n$$ has at least two elements. But then $$a = (a_n)_n$$ is an element of $$\prod_{n \in \mathbb{N}} A_n$$ which satisfies $$a \neq f(n)$$ for all $$n \in \mathbb{N}$$ and so $$f$$ is not surjective.