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Frankly, I have a hard time imagining what it is.

In his Cours d'arithmétique(1970), Jean-Pierre Serre defines $$\forall n\geq 1, A_n:=\mathbb Z/3^n\mathbb Z$$

Let us consider $A_1$. Serre's talking about "discrete topology on $A_1$", without even defining it. I've been looking: if I'm not mistaken, this is the topology defined by the set of parts of $A_1$ $$\tau_1:=\{ \emptyset, \{0\},\{1\},\{2\},\{0,1\},\{0,2\},\{1,2\},\{0,1,2\} \}$$ It is clear that $$\forall n\geq1(A_n,\tau_n) \text{ is separated space}$$ So, every $A_n $ is compact (finited unions of points) . Then he writes :" If we give $\prod_{}^{}A_n$ the product topology , $Z_3$ has a topology that makes it a compact space (because it is closed in a compact space)."

I know that a closed space in a compact space is compact. The product topology $\tau$ is defined, I believe, by the set of sets of the form $$O_1\times O_2\times...\times O_m\times A_{m+1}\times A_{m+2\times...}$$

1.- Why $\prod_{}^{}A_n$ is compact ?

2.- Why $\mathbb Z_3$ is closed in $\prod_{}^{}A_n$ ?

Thank you in advance for your help.

Stéphane Jaouen
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2 Answers2

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Serre's talking about "discrete topology on $A_1$", without even defining it.

Discrete topology on set $X$ is always given by full $\mathcal{P}(X)$, i.e. the collection of all subsets. Meaning every subset is open. This is a well known definition.

1.- Why $\prod_{}^{}A_n$ is compact ?

Because arbitrary product of compact spaces is compact. This is known as Tychonoff's theorem.

2.- Why $\mathbb Z_3$ is closed in $\prod_{}^{}A_n$ ?

$\mathbb{Z}_3$ in this context is defined as the inverse limit. Given quotient maps $\pi_{i,j}:\mathbb{Z}/3^j\mathbb{Z}\to \mathbb{Z}/3^i\mathbb{Z}$ for $i\leq j$ we define

$$\mathbb{Z}_3=\{(a_n)\in \prod A_n\ |\ a_i=\pi_{i,j}(a_j)\ |\ i\leq j\}$$

In general an inverse limit does not have to be closed in the product space. But it is always closed, if each $A_k$ space is Hausdorff, as you can see here: Is inverse limit topology closed in product topology? And in our case each space is discrete, and thus Hausdorff.

freakish
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  • Thank you. Unfortunately, I don't have Tychonov's theorem in Dixmier's General Topology but I have it in Skandalis Topology and analysis,3rd year; p.290 all the same:( And the link to explain my question 2) is cool. – Stéphane Jaouen Apr 04 '24 at 07:12
  • Tikhonov's theorem applies to families of general compact topological spaces, however complicated they may be. Here, these are very simple compact spaces. Wouldn't there be a way to prove the result without referring to Tychonov's theorem in this simple case? – Stéphane Jaouen Apr 04 '24 at 08:33
  • Maybe, but I think the interesting case of Tychonoff's theorem is when the product is infinite, as it is here. (It is elementary that a finite product of compact spaces is compact.) The $A_n$ are trivially compact because finite, but their product is not. – MJD Apr 04 '24 at 08:42
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    Also note that an infinite product of discrete spaces is not discrete. – MJD Apr 04 '24 at 08:48
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    @StéphaneJaouen have a look at this: https://math.stackexchange.com/questions/4466504/countable-product-of-finite-topological-spaces-is-compact-without-tychonoffs-t – freakish Apr 04 '24 at 08:49
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  1. The link given by @freakish is perfect.
  2. I found another answer that seems useful to me. Let $\psi_n:\prod_k A_k\to A_{n-1} $ defined by $$\psi_n(x)=\varphi_n(x_n)-x_{n-1}$$ Then, $\mathbb Z_3=\cap_n\psi_n^{-1}(\{0\})$ is an intersection of closed subsets :$\psi_n^{-1}(\{0\})$ is closed because it is the reciprocal image of a closed subset by a continuous function.

So, $\mathbb Z_3$ is closed.$\square$

Note that $$\boxed{\mathbb Z_3:=\varprojlim (A_n,\varphi_n)}$$ to know what $\varphi_n $ is(see also here.)

Stéphane Jaouen
  • 3,625