Problem 1

Exercise 9.1 (Product and box topologies)   Let

$$X = \prod_{n=1}^\infty [0,1]$$ (9.1)

$$S = \{ (x_n) \in X \;\vert\; \exists N : n\geq N \implies x_n=0 \}$$ (9.2)

$$= \bigcup_{N=1}^\infty \{ (x_n) \;\vert\; n\geq N \implies x_n=0 \}$$ (9.3)

1. Show that if $X$ is considered with the product topology, then the closure of $S$ is $X$.
2. Show that if $X$ is considered with the box topology, then $S$ is closed in $X$.

Proof. For part (a), Let $x=(x_n)\in X\setminus S$. We will show $x$ is a limit point of $S$. Define a sequence of elements in $S$:

$$s^1 := (x_1, 0, \dots)$$ (9.4)

$$s^2 := (x_2, x_2, 0, \dots)$$ (9.5)

$$\quad \vdots$$ (9.6)

$$s^n := (x_1, x_2, \dots, x_n, 0, 0,\dots)$$ (9.7)

If $U$ is a given neighborhood of $x$, we argue that $s^n\in U$ for some $n$. From the definition of the product topology, select a basic element $B\subseteq U$ containing $x$ and realize $B$ as a product of finitely many not necessarily trivial sets:

$$B = \left(\prod_{n=1}^N U_n\right)\times\prod_{n=N+1}^\infty [0,1]$$ (9.8)

Then $s^N\in B\subseteq U$.

For part (b), suppose $x\in X\setminus S$ is a limit point of $S$. Select

$$s \in \prod_{n=1}^\infty \begin{cases} B_{\vert x_n\vert}(x_n) \quad x_n\neq 0\\ B_1(0) \quad x_n = 0 \end{cases}$$ (9.9)

Since $x\notin S$, there exists a subsequence $x_{n_k}$ such that $x_{n_k}\neq 0$, so we may write the product as

$$\left(\prod_{k=1}^\infty B_{\vert x_{n_k}\vert}(x_{n_k})\right)\times \prod_{x_n=0}B_1(0)$$ (9.10)

We can see that if $s$ lies in the first product, then $s$ does not lie in $S$, because $0\notin B_{\vert x_{n_k}\vert}(x_{n_k})$, so there is no $N$ after which all the elements are zero. This is a contradiction, so we know $x\in S$. Therefore, $S$ is closed. $\qedsymbol$