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Let $J$ be an arbitrary index set. Then how to prove that the uniform topology on the Cartesian product $\mathbf{R}^J$ of the set $\mathbf{R}$ of real numbers with itself is different from both the product and the box topologies when $J$ is infinite?

It's already known to me how the uniform topology is finer than the product topology and coarser than the box topology.

The uniform metric $d\colon \mathbf{R}^J \times \mathbf{R}^J \to \mathbf{R}$ is given by $$ d(x,y) := \sup_{\alpha\in J} \min(1, |x_{\alpha}-y_{\alpha}|) $$ for any two points $x := (x_{\alpha})_{\alpha\in J}$, $\,$ $y := (y_{\alpha})_{\alpha\in J}$ $\,$ in $\mathbf{R}^J$.

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2 Answers 2

I'll just work with $J = \mathbb{N}$. Similar examples can be made for all $J$ by focusing on some countably infinite subset of $J$.

Hint 1: The set $$U = \{ x = ( x_n )_{n \in \mathbb{N}} : ( \forall n \in \mathbb{N} ) ( | x_n | < 2^{-n} ) \}$$ is open in the box topology, but not in the uniform topology.

Hint 2: The set $$V = \{ x = ( x_n )_{n \in \mathbb{N}} : ( \forall n \in \mathbb{N} ) ( | x_n | < \tfrac 12 ) \}$$ is open in the uniform topology, but not in the product topology.

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How can we explicitly demonstrate that the set $U$ is not open in the uniform topology? –  Saaqib Mahmuud Feb 7 '13 at 16:04
@SaaqibMahmuud: You should be able to show that no (uniform) ball is a subset of $U$. (At the very least, it should be easy to show that no ball centred at the constant zero sequence is a subset of $U$.) –  Arthur Fischer Feb 7 '13 at 16:20
Goodness me! It was easy. Thank you so much for being so helpful. Wow! You seem to know it all! –  Saaqib Mahmuud Feb 7 '13 at 16:26

Here is an example to demonstrate that the uniform topology is different from the box topology when $J = \omega$. It is from the exercise 6 of section 20 titled "The Metric Topology" of Munkres's Topology.

Given $\mathbf{x} = (x_1, x_2, \ldots) \in \mathbb{R}^{\omega}$ and given $0 < \epsilon < 1$, let $$U(\mathbf{x,\epsilon}) = (x_1 - \epsilon, x_1 + \epsilon) \times \cdots \times (x_n - \epsilon, x_n + \epsilon) \times \cdots.$$

$U(\mathbf{x}, \epsilon)$ is open in box product, but not in the uniform topology.

For the reason why $U(\mathbf{x}, \epsilon)$ is not open in the uniform topology, consider the particular point $(x_1 + \epsilon/2, x_2 + 2\epsilon/3, x_3 + 3\epsilon/4, \cdots)$ which is obviously in $U(\mathbf{x}, \epsilon)$. However, there is no ball (in the uniform topology) centered around it that is in $U(\mathbf{x}, \epsilon)$.

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