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Let $A$ be some finite alphabet. Let $A$ be equipped with the discrete topology and $A^{\mathbb{Z}}$ equipped with the associated product topology.

Am I right that each function

$f\colon A^{\mathbb{Z}}\to A^{\mathbb{Z}}$ is continuous?

H. Hawks
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2 Answers2

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Let $A=\left\{a,b\right\}$. Consider the map $f:A^{\mathbb{Z}}\rightarrow A^{\mathbb{Z}}$ that maps $f(\prod_{-\infty}^{\infty}a)= \prod_{-\infty}^{\infty}a$ and anything else to $\prod_{-\infty}^{\infty}b$. Then $f^{-1}(\prod_{i=-\infty}^{\infty}X_i)= \prod_{-\infty}^{\infty}\left\{a\right\}$ where $X_i=A$ for all $i\neq 0$ and $X_0=\left\{a\right\}$. Since $\prod_{i=-\infty}^{\infty}X_i$ is open in $A^{\mathbb{Z}}$ and $\prod_{-\infty}^{\infty}\left\{a\right\}$ is not, $f$ is not continuous. Using this argument you can easily see that $f$ need not be continuous as soon as $|A|\geq 2$.

Mathematician 42
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  • What do you mean with $\prod_{-\infty}^{\infty}a$? – H. Hawks May 13 '16 at 13:18
  • An element of $A^\mathbb{Z}$ under the identification $A^\mathbb{Z}$ with $\prod_\mathbb{Z}{A}$ – cQQkie May 13 '16 at 13:21
  • Then I do not understand the description of the function. An element is imaged to itself and anything else to another element? – H. Hawks May 13 '16 at 13:25
  • Sry, I should add: it means the string with constant $a$, ie the function $\mathbb{Z}\to A$ picking out $a$ for every $z\in\mathbb{Z}$ – cQQkie May 13 '16 at 13:26
  • A function from $g:\mathbb{Z}\rightarrow A$ can be represented as a two-sided sequence. Hence $g$ can be written as $g(i){i\in \mathbb{Z}}$ or equivalently $\prod{i=-\infty}^{\infty} g(i)$. Thus $\prod_{-\infty}^{\infty} a$ denotes the constant sequence $a$. Any other sequence is mapped to the constant sequence $b$. – Mathematician 42 May 13 '16 at 13:29
  • Okay, but I do not see (1.) Why $f^{-1}(\prod_{i=-\infty}^{\infty}X_i)$ is $\prod_{i=-\infty}^{\infty}a$ and, moreover, why is this pre-image not open? – H. Hawks May 13 '16 at 13:31
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    Since $X_0=\left{a\right}$ we need sequences that are mapped to something with an $a$, but there is only one such sequence, namely the constant sequence $a$ (since all other sequences are mapped to the constant sequence $b$). The preimage is not open since it is an infinite product of the opens $\left{a\right}$, this is not open since by definition of the product topology we only allow finitely many opens different from $A$ itself. – Mathematician 42 May 13 '16 at 13:41
  • Consider the cylinder sets $[a_m,\ldots,a_n]:=\left{x\in A^{\mathbb{Z}}: x_j=a_j~\forall m\leq j\leq n\right}$ with $a_j\in A, m\leq n$. These sets form a base of the product topology and are clopen. I think the cylinder sets also form a base for the closed subsets of $X=A^{\mathbb{Z}}$. - To show that a function $f\colon X\to X$ is continuous, it should be enough to show that the pre-image of a cylinder set is closed, i.e. can be expressed as an intersection of cylinder sets. – H. Hawks May 13 '16 at 17:41
  • Now, consider any cylinder set for your example. For example $[a]0$. This has pre-image $\left{a\right}^{\mathbb{Z}}$. But why not writing this as $\bigcap{m\geq 1}[a_{-m},\ldots,a_m]$ with $[a_{-m},\ldots,a_m]=(a,a,\ldots,a)$ for each $m\geq 1$? So isn't it true that the pre-image of each cylinder set is closed? . – H. Hawks May 13 '16 at 17:43
  • No, this is not true. You showed that the preimage of $[a]_0$ is closed, but you didn't show that the preimage of any closed is closed (which you need to do if you want to argue that $f$ is continuous). – Mathematician 42 May 13 '16 at 23:10
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Note that $X = A^{\mathbb Z}$ is Hausdorff but not discrete. So there is a point $p \in X$ such that the singleton $\{p\}$ is not an open set. Choose a second point $q \ne p$. Define $f : X \to X$ by $$ f(p) = q,\qquad\text{and}\qquad f(x) = p \text{ for all $x \ne p$}. $$ Then $U = X \setminus \{p\}$ is open, but $f^{-1}(U) = \{p\}$ is not open.

GEdgar
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  • Is it really Hausdorff? – H. Hawks May 13 '16 at 14:28
  • Why not try to prove that yourself? – GEdgar May 13 '16 at 14:42
  • As far as I know, the cylinder sets $[a_{-n},\ldots,a_n]$ generate the product- topology. Moreover, A^Z is compact and as you say, Hausdorff. Then the pre-image of a cylinder set (which are closed and open), should be a subset of a cylinder set and hence closed since a cylinder set is compact (?). So up to this the function is continuous.... thats why i am confused to hear that not each function f is continuous. – H. Hawks May 13 '16 at 14:46
  • ??? a subset of a cylinder set is closed ??? I doubt it. – GEdgar May 13 '16 at 14:48
  • Please see the two last comments which I wrote below the other answer. – H. Hawks May 13 '16 at 17:45