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Let $A = \{a_1,\dots,a_m\}$ be a finite set endowed with a discrete topology and let $X = A^{\Bbb N}$ be the product topological space. I wonder which bounded functions $f:X\to\Bbb R$ are continuous on $X$.

For example, it is clear that if $f$ depends only on a finite number of coordinates then $f\in C(X)$, i.e. if there exists some finite $n$ such that $$ f(x_1,\dots,x_n,x_{n+1},x_{n+2},\dots) = f(x_1,\dots,x_n,x'_{n+1},x'_{n+2},\dots) \quad \forall x_{n+1},x_{n+2},x'_{n+1},x'_{n+2},\dots $$ then $f\in C(X)$. Thus it seems that only dependents on infinitely many coordinates may violate continuity. I would be happy, if one could tell me whether there are some useful necessary/sufficient conditions to assure $f\in C(X)$.

In particular, if $B\subset A$ and $1_B(a)$ is the indicator (characteristic) function of $B$, does it hold that $$ g(x):=\limsup\limits_{k\to\infty}1_B(x_k) $$ is a continuous function on $X$.

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@tohecz: thanks, that was a typo - fixed now –  S.D. Nov 28 '12 at 10:38
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Assuming that $m>1$, the space $X$ is just a Cantor set: up to homeomorphism it’s the same space for all $m>1$, and you’re really asking which bounded, real-valued functions on the Cantor set are continuous. –  Brian M. Scott Nov 28 '12 at 11:21
    
@BrianM.Scott: thanks for the comment - so does the characterization by toehz hold true? –  S.D. Nov 29 '12 at 7:55
    
It’s correct, but I don’t yet see how to convert it into a nice, easily tested criterion. –  Brian M. Scott Nov 29 '12 at 9:03
    
@BrianM.Scott: well, with focus on the indicator functions it yields some nice characterization that I've mentioned in my first comment to tohecz's answer below (hopefully, this comment is correct) –  S.D. Nov 29 '12 at 12:32

2 Answers 2

up vote 3 down vote accepted

Yes, there are, and probably numerous. The basic idea is the following:

Theorem.

Let $[a_1\dotsm a_n]:=a_1\dotsm a_n A^{\mathbb N}$ denote the set called a cylinder. Then basically, $f$ is continuous if an only if $$ \operatorname{diam} f\bigl([a_1\dotsm a_n]\bigr)\xrightarrow{n\to\infty}0 \quad\text{for all}\quad a_1a_2a_3\dotsm\in A^{\mathbb N},$$ where $\operatorname{diam}Y$ is the diameter of a set.

This property is crucial for all numeration systems. For example, for the decimal expansion one has $f(a_1a_2a_3\dotsm)=\frac{a_1}{10}+\frac{a_2}{100}+\frac{a_3}{1000}+\dotsb$ and diameter of image of cylinder of $[a_1\dotsm a_n]$ is exactly $10^{-n}\to0$, hence $f$ is continuous. The same is true for the binary system as well, generally for all radix representations. But it is true even for continued fractions when written as an infinite seqence of matrices $L=\begin{pmatrix}1&0\\1&1\end{pmatrix}$, $R=\begin{pmatrix}1&1\\0&1\end{pmatrix}$, and for many other systems as well.

Proof.

On the set $A^{\mathbb N}$ we use the (sometimes called) Cantor topology given by a Cantor metric: $\rho(\mathbf{a},\mathbf{b})=2^{-k}$ where $k=\min\{j:a_j\neq b_j\}$ (bold letters indicate infinite words).

Continuity of $f$ means $$ (\forall \mathbf x\in A^{\mathbb N})(\forall\varepsilon>0)(\exists\delta>0)(\forall \mathbf y\in A^{\mathbb N}, \rho(\mathbf x,\mathbf y)<\delta)(\lvert f(\mathbf y)-f(\mathbf x)\rvert<\epsilon)$$

Since the sets of such $\mathbf y$ are exactly some cylinders, this should be obviously equivalent to the theorem statement.

Remark.

Notice that all the cylinders are clopen sets, which is very interesting itself.

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Thanks a lot, that's a very nice characterization. It means for example, that whenever $f = 1_Y$ is an indicator function with $Y\subset X$, its image has diameter either $0$ or $1$. Thus, for it to be continuous, it is equivalent to the following of property: for any $x\in X$ the belonging of $x$ to $Y$ has to be decided only on a finite number of first coordinates. Awesome! Could I ask you on the way how to prove the characterization via the diameter of the image of a cylinder set you've provided? –  S.D. Nov 28 '12 at 10:43
    
@S.D. It depends what exactly you mean by "product topology". I use the "Cantor distance" of two strings: Let $a_1a_2a_3\dotsm$, $b_1b_2b_3\dotsm$. Let $\ell:=\min\{j:a_j\neq b_j\}$. Then $\rho(\mathbf a,\mathbf b)=2^{-j}$. Tell me if this corresponds to your topology. If so, I'll provide the proof. –  tohecz Nov 28 '12 at 10:54
    
I use the standard definition as the smallest topology w.r.t. which projections are continuous, see e.g. here. I didn't do topology for some time, so I am not sure whether it is metrizable in case I am talking about. Maybe, it is. It would be nice, though, if you would have a pure topological proof for the fact that you've mentioned. –  S.D. Nov 28 '12 at 11:00
    
@S.D.: You’re both talking about the same topology; tohecz’s cylinders form a base for the product topology. –  Brian M. Scott Nov 28 '12 at 11:20
    
@tohecz: as Brian have mentioned it does correspond to my topology - would you please show the proof? –  S.D. Nov 30 '12 at 9:34

I do not know whether there are general conditions, but I can answer your particular question: Note that the sequence $(y^n)$ with $y^n = (x_1, x_1, \ldots, x_1, x_2, \ldots)$ with $n$ $x_1$'s converges to $y = (x_1, x_1, \ldots)$ in $A^{\mathbb N}$. Now let $B = \{x_2\}$, we have for any $n$: $$ g_B(y^n) = \limsup_{k\to\infty} 1_B(y^n_k) = 1 $$ (as $y^n_k = x_2$ for $k> n$). But $$ g_B(y) = \limsup_{k\to\infty} 1_B(x_1) = 0 $$ So $y^n \to y$, but $g(y^n) \not\to g(y)$ hence $g$ isn't continuous.

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thanks a lot for the answer –  S.D. Nov 29 '12 at 12:33

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