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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?

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    $\begingroup$ what is this $Z$? $\endgroup$ – Alvin Lepik Jul 30 '19 at 8:53
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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.

We did not really use that we are dealing with the integers by the way. We only need a set large enough to choose different elements, i.e. containing at least two elements.

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The infinite Cartesian product of non-empty, non-singleton sets is uncountable infinite.

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.

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