A non-empty finite set of symbols, such as {$0, 1$}, is called an alphabet. A word over an alphabet $Σ$ is a tuple $w = (w_1, w_2, . . . , w_n)$ $($often written simply as $w_1w_2 · · · w_n)$ where $n ≥ 0$ and $w_1, w_2, . . . , w_n ∈ Σ$. The
number $n$ is called the length of $w$.
The unique word with length $0$ is denoted by $λ$. The set of all words over
an alphabet $Σ$ is denoted by $Σ
^∗$
, thus $Σ^∗:= U_{n≥0} Σ^n$.
An infinite sequence $x = (x_1, x_2, . . .)$ with elements from
an alphabet $Σ$ is called a (one-sided) configuration. Again, a configuration $(x_1, x_2, . . .)$ is often written more
concisely as $x_1x_2 · · ·$ .
As an example, $01101$ is a word of length $5$ over the alphabet {$0, 1$}, and $1101001000100001 · · · $ is a configuration over the same alphabet.
Let $X := Σ^{\mathbb N} $ be set of all configurations over an alphabet $Σ$. Every word $w = w_1w_2 · · · w_n ∈ Σ
^∗$ defines a
subset of $X$ by
$C(w) :=$ {$x ∈ X : x_1x_2 · · · x_n = w_1w_2 · · · w_n$}.
Consider the space $X := Σ^\mathbb N$ of all configurations over an alphabet $Σ$.
Prove that the product topology on $X$ (where $Σ$ is given the discrete topology) coincides with the topology
generated by the cylinder sets.
My attempt:
Let $C(w) :=$ {$x \in X: x_1x_2…x_n = w_1w_2…w_n$} be the set of cylinders which is a basis for the topology on $X$. (I have already proven this).
Let $T_1 :=$ product topology on X.
Let $T_2 :=$ topology generated by the cylinder sets.
I need to prove that $T_1 = T_2$, i.e.:
$1) T_1 \subset T_2$
$1) T_2 \subset T_1$
So, $1) T_1 = T(Σ^\mathbb N) =$ all unions of configurations $=$ all unions of elements of $C(w) ($as $C(w)$ is a subset of $X$ and $C(\lambda) = X$, where $\lambda$ is the unique word of length $0) \subset T_2$.
$2) T_2 = T(C(w)) =$ coarsest topology containing $C(w) => T_2 \subset T_1$
Hence $T_1 = T_2$
Is my attempt correct? Any other answers?
Best Answer
Let $\mathcal{T}_c$ be the topology generated by the collection of all cilinder sets $\{C(w)\mid w \in \Sigma^\ast\}$ as a base.
Let $\mathcal{T}_p$ be the product topology on $X=\Sigma^{\Bbb N}$ ($\Sigma$ discrete), which is by definition the smallest topology that makes all projections continuous, where a projection $\pi_n$, $n \in \Bbb N$, sends $x_1x_2\ldots \in X$ to $x_n$, its $n$-th component/element.
By $l(w)$ I'll denote the length of a word $w \in \Sigma^\ast$.
Let $\sigma \in \Sigma$ be any symbol from the alphabet. Then $$\pi_n^{-1}[\{\sigma\}] = \{x \in X\mid x_n = \sigma\} = \bigcup\{C(w)\mid w \in \Sigma^\ast: l(w)=n \text{ and } w_n = \sigma \}$$
which shows that the inverse image of a singleton under any projection is open in $\mathcal{T}_c$. As all open subsets of $\Sigma$ are just unions of singletons (they form a base), it follows that $\mathcal{T}_c$ is a topology that makes all $\pi_n$ continuous, and so by minimality of the product topology: $$\mathcal{T}_p \subseteq \mathcal{T}_c$$
On the other hand, if $w \in \Sigma^\ast$ is some word:
$$C(w) = \bigcap_{i=1}^{l(w)} \pi_i^{-1}[\{w_i\}]$$
which shows that all $C(w) \in \mathcal{T}_p$ (finite intersection of sets that are open by continuity of the projections under $\mathcal{T}_p$) and so it follows that also
$$\mathcal{T}_c \subseteq \mathcal{T}_p$$
and equality of topologies ensues.