# Probs. 2 (b) and 2 (c), Sec. 25 in Munkres' TOPOLOGY, 2nd ed: Components in the uniform and box topologies

Let $X$ be a topological space. Then, for any points $x, y \in X$, we can define a relation $\sim$ on $X$ by defining $x \sim y$ iff there is a connected subspace of $X$ containing both $x$ and $y$. This relation is an equivalence relation on $X$, and the equivalence classes determined by it are called the components (or connected components) of $X$.

Let $\mathbb{R}^\omega$ denote the set of all the (infinite) sequences of real numbers.

Problem 2 (b), Sec. 25 in Munkres: Let $\mathbb{R}^\omega$ have the uniform topology. Then how to show that $x$ and $y$ are in the same component of $\mathbb{R}^\omega$ if and only if the sequence $$x-y = (x_1 - y_1, x_2 - y_2, x_3 - y_3, \ldots)$$ is bounded?

Problem 2 (c), Sec. 25 in Munkres: Let $\mathbb{R}^\omega$ have the box topology. Then how to show that $x$ and $y$ are in the same component of $\mathbb{R}^\omega$ if and only if the sequence $$x-y = (x_1 - y_1, x_2 - y_2, x_3 - y_3, \ldots)$$ is eventually zero?

## 1 Answer

First part: Let $\mathbb{R}^{\omega}$ have the uniform topology, and let $x,y \in \mathbb{R}^{\omega}$. Let $||\cdot||_u$ denote the uniform (pseudo)metric on $\mathbb{R}^{\omega}$, i.e, $||x||_u = \sup_n |x_n|$

Case $1$: Suppose $x-y$ is bounded, and define $f: [0,1] \to \mathbb{R}^{\omega}$ as $t \mapsto x+t(y-x)$. Then $f$ is continuous (in fact, Lipschitzian) because $||f(s)-f(t)||_u = |s-t|\cdot ||x-y||_u$. Since $[0,1]$ is connected and $f$ is continuous, it follows that $f([0,1])$ is a connected subset of $\mathbb{R}^{\omega}$ which contains $x$ and $y$. Therefore $x \sim y$.

Case $2$: Suppose $x-y$ is unbounded. Then form a separation of $\mathbb{R}^{\omega}$ as follows: let $U = \{ z \in \mathbb{R}^{\omega}: x-z$ is bounded $\}$ and let $V = \mathbb{R}^{\omega} \backslash U$. Then $U$ and $V$ are open and disjoint (it is easy to check), and their union is $\mathbb{R}^{\omega}$. Also $x \in U$ and $y \in V$, so $x$ and $y$ cannot be in the same connected component.

Second Part: Let $\mathbb{R}^{\omega}$ have the box topology, and let $x,y \in \mathbb{R}^{\omega}$.

Case $1$: Suppose $x_n=y_n$ for $n > N$. Define a map $f: \mathbb{R}^N \to \mathbb{R}^{\omega}$ as $(a_1,...,a_N) \mapsto (a_1,...,a_N,x_{N+1},x_{N+2},...)$. Then $f$ is continuous, since $f^{-1}(\Pi_{n\in\mathbb{N}}U_n) = \Pi_{n=1}^N U_n$ as long as $x_n \in U_n$ for $n > N$ (otherwise $f^{-1}(\Pi_{n\in\mathbb{N}}U_n) = \emptyset$). Since $\mathbb{R}^N$ is connected, it follows that $f\left(\mathbb{R}^N\right)$ is connected. Since $x,y \in f\left(\mathbb{R}^N \right)$, it follows that $x \sim y$.

Case $2$: Suppose that $x_n \neq y_n$ for infinitely many $n$. Define a map $g: \mathbb{R}^{\omega} \to \mathbb{R}^{\omega}$ as $$(g(z))_n = \begin{cases} \frac{nz_n}{x_n-y_n} & \text{if } x_n \neq y_n \\ z_n & \text{if } x_n=y_n \end{cases}$$ Then $g$ is continuous with respect to the box topology, because for any collection $\{U_n\}_{n\in\mathbb{N}}$ of open subsets of $\mathbb{R}$, the set $g^{-1}(\Pi_{n\in\mathbb{N}}U_n)$ is easily checked to be a set of the form $\Pi_{n\in\mathbb{N}} V_n$ for some open sets $V_n \subset \mathbb{R}$ (explicitly, $V_n = U_n$ if $x_n=y_n$, and $V_n = \frac{x_n-y_n}{n} \cdot U_n$ if $x_n \neq y_n$). It is easily checked that $g(x)-g(y)=g(x-y)$ is an unbounded sequence (since $(g(x-y))_n = n$ whenever $x_n \neq y_n$). Therefore, using the first part of the question (and the fact that the box topology contains the uniform topology), there exist open sets $U,V$ which form a separation of $\mathbb{R}^{\omega}$ and $g(x)\in U$ and $g(y) \in V$. Thus $g^{-1}(U)$ and $g^{-1}(V)$ form a separation of $\mathbb{R}^{\omega}$, with $x \in g^{-1}(U)$ and $y\in g^{-1}(V)$. So $x$ and $y$ are not equivalent.

• @SaaqibMahmuud: I have been trying to think of an easier argument for Case 2 in the second part. It appears that explicitly, we can let $U = \{z \in \mathbb{R}^{\omega} : \big|\frac{n(x_n-z_n)}{x_n-y_n}\big|$ is bounded $\}$, and let $V = \mathbb{R}^{\omega} \backslash U$. But a direct proof that $U$ is open would not be so simple... – Shalop May 9 '15 at 18:00
• I wonder if you've made some typo or other mistake in Case 2 of the second part, for I'm unable to fully grasp it. So can you please go over it again and see if there's any corrections worth making or any details worth incorporating so as to make it more easily understandable. – Saaqib Mahmood May 9 '15 at 18:37
• thank you for your wonderful solution! However, in part (b), I would prefer you to directly prove the required assertions by using the uniform metric $\bar{\rho}$ on $\mathbb{R}^\omega$, which is defined as follows: $$\bar{\rho} ( \mathbf{x}, \mathbf{y})\colon=\sup\left\{\ \min\left\{\ \left\lvert x_n-y_n\right\rvert, 1 \ \right\}\ \colon \ n\in\mathbb{N}\ \right\}$$ for all $\mathbf{x}\colon=\left(x_n\right)_{n\in\mathbb{N}}$ and $\mathbf{y}\colon=\left(y_n\right)_{n\in\mathbb{N}}$ in $\mathbb{R}^\omega$. Or, please prove the assertions by using the corresponding "uniform" norm. – Saaqib Mahmood Jul 10 '18 at 8:51