# Topology on the set of infinite sequences

Help, who can give me an example to show the following definition of $\lim$ and closed set

let $\Sigma$ be a set of states, let $\Sigma^\omega$ denote the set of all infinite sequences of elements in $\Sigma$. For a sequence $\langle \langle \sigma_0, \sigma_1, \sigma_2, ...\rangle\rangle$,let $\sigma|_m$ denote $\langle\langle s_0,s_1,...s_{m-1}\rangle\rangle$, the prefix of $\sigma$ of length $m$. An infinite sequence $\langle \langle \sigma_0, \sigma_1, \sigma_2, ...\rangle\rangle$ of sequences in $\Sigma^\omega$ is said to converge to the sequence $\sigma$ in $\Sigma^\omega$ iff for all $m\geq0$, there exists an $n\geq 0$ such that $\sigma_i|_m=\sigma|_m$ for all $i\geq n$. In this case, we define $\lim\ \sigma_i$ to be $\sigma$. Let $\sigma$ be an element of $\sigma^\omega$ and let $S$ be a subset of $\sigma^\omega$. We say that $\sigma$ is a limit point of $S$ iff there exist elements $\sigma_i$ in $S$ such that $\lim \ \sigma_i = \sigma$. The set $S$ is closed iff $S$ contains all its limit points. The closure of $S$, denoted $\overline{S}$, consists of all limit points of $S$; it is the smallest closed superset of $S$.

## migrated from mathoverflow.netMar 15 '14 at 10:38

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• I think $\sigma^w$ should be $\Sigma^w$ here. Also, this is neither algebraic topology nor research mathematics. – Michael Greinecker Mar 15 '14 at 9:53

## 2 Answers

I will assume that you know what a metric space is and have worked with metric spaces before. Given this assumption, I will define a metric on the space $\Sigma^{\omega}$ which coincides with the definition of limit and closure which you have given (and is the usual metric/topology we place on $\Sigma^{\omega}$).

Let $\sigma=\langle s_0,s_1,s_2,\ldots\rangle$, $\sigma'=\langle s_0',s_1',s_2',\ldots\rangle$ be in $\Sigma^{\omega}$ and for $a,b\in\Sigma$ let $\delta(a,b)=0$ if $a=b$ and $\delta(a,b)=1$ if $a\neq b$ (this is the kronecker delta function on the symbols $\Sigma$). Finally, we define the metric $d$ on $\Sigma^{\omega}$ by $$d(\sigma,\sigma')=\sum_{i=0}^{\infty}2^{-i}\delta(s_i,s_i').$$

It's a fairly simple exercise to show that $d$ is a metric and that the open balls in $(\Sigma^{\omega},d)$ contain elements which either agree on a large block of symbols from the first symbol, or else agree on a larger block of symbols after 'shifting' both of the symbols to the left a small amount and deleting the left-most symbols as we shift (the more we shift, the larger the block they will have to agree on).

As a very simple example of a sequence of elements of $\Sigma^{\omega}$ which converges to some limit, take the following in the space $\{0,1\}^{\omega}$. $$\sigma_0=\langle 0,0,0,0\ldots\rangle$$

$$\sigma_1=\langle 1,0,0,0\ldots\rangle$$

$$\sigma_2=\langle 1,1,0,0\ldots\rangle$$

$$\sigma_3=\langle 1,1,1,0\ldots\rangle$$

and so $\sigma_n$ starts with $n$ $1$s and then has a tail of repeating $0$s. I claim this sequence of symbol sequences has a limit $\sigma$ which is equal to $$\sigma=\langle 1,1,1,1,1,1,... \ldots \rangle$$ and indeed $d(\sigma_n,\sigma) = \sum_{i=0}^{\infty} 2^{-i} d((s_n)_i,1)$ and so by how we have defined $\sigma_n$, we know that $(s_n)_i=1$ if and only if $n\leq i$. It follows that $d(\sigma_n,\sigma) = \sum_{i=n}^{\infty} 2^{-i}=2-\sum_{i=0}^{n} 2^{-i}$. We note then that $$\lim_{n\to\infty} d(\sigma_n,\sigma)=2-\lim_{n\to\infty} \sum_{i=0}^{n} 2^{-i}=2-2=0$$ and so conclude that indeed $\lim_{n\to\infty}\sigma_n=\sigma$ as claimed.

This notion of limit in the metric space sense coincides with the definition that you have been given.

Given the above, I'll leave the closed set example to you. Just note that again it is the same as a closed set in the above metric space.

Take $\Sigma=\{0,1,2,3,4,5,6,7,8,9\}$. You can define the function $f:\Sigma^\omega\longrightarrow (0,1)$ that sends a sequence $\sigma\in\Sigma^\omega$ to the number that (in base $10$) begins with $0.$ and its digits are the terms of $\sigma$, in the same order. This function is continuous, so you can think in it for an example.