Continuity of a map between a product space equipped with product topology

Question

Consider $[n]=\{1, 2, \cdots, n\}$ with the discrete topology and let $X=\prod_{n\geq 1}[n]$ be the product space with the product topology. For $x=(a_1, a_2, \cdots)$, define $T(x)=(1, a_1, a_2, \cdots)$. Then how to show the following?

1. The map $T:X\to X$ has a unique fixed point

2. The map $T:X\to X$ is continuous

For part 1, let $T(x)=x$, then $(a_1, a_2, \cdots)=T(x)=(1, a_1, a_2, \cdots)$ which means $a_1=1$, $a_2=a_1=1$, $a_3=a_2=1, \cdots$. Therefore, $x=(1, 1, 1, \cdots)$ is the unique fixed point.

But I don't know how to prove the second part. Please help how to initiate? Thanks in advance.

Answer 1

A standard fact on the product topology is applicable here:

$f: Y \to \prod_i X_i$ is continuous iff $\pi_i \circ f$ is continuous for all $i$, where the $\pi_i$ are the standard projection maps.

All this is classical (and characterises the product topology), e.g. see my post here on initial topologies in general.

For your case, we use $Y=X$ and the product $\prod_{n \ge 1} [n]$ as our product space. $\pi_1 \circ T$ is constantly $1$, so continuous, and for $n \ge 2$ we have $$\pi_n \circ T = \pi_{n-1}$$ which is also continuous, so the above criterion tells us that $T$ is continuous.

Answer 2

Consider $U$ a basic open subset of the product space. That means that $U$ is the product of open subsets of the spaces $[n]$, from which though, only finite of them are not equal to the respective whole space $[n]$. Observe that a product $[1]=\{1\}$ so every open set in the product topology is of the form $\{1\}\times U_2 \times U_3 \times \cdots$, where $U_n$ is on open set in $[n]$.

Now also observe that $T(a_1, a_2,\ldots)=(1,1,a_2,...)$. Thus if $U_2$ is an open set that doesn't contain $1$, i.e. $U_2=\{2\}$, then the inverse image of $U$ is the empty set.

If $U_2$ contains $1$, the inverse image of $U$ is the set $V=\{1\} \times U_3 \times U_4\times \cdots$.

In both cases the inverse image is an open set.