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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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I'll just work with $J = \mathbb{N}$. Similar examples can be made for all infinite $J$.

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

    As $U = \prod_{n \in \mathbb{N}} ( -2^{-n} , 2^{-n} )$, it is a product of open intervals, so it is clearly open in the box topology. It is not open in the uniform topology since it doesn't include any uniform-ball centred at $\mathbf{0}$. (Given $\varepsilon > 0$ take $k$ such that $2^{-k} < \varepsilon$ and note that the point $\mathbf{y} = ( 2^{-k} )_{n \in \mathbb{N}}$ belongs to the uniform $\varepsilon$-ball centred at $\mathbf{0}$, but does not belong to $U$.)

  • The set $$V = \{ \mathbf{x} = ( x_n )_{n \in \mathbb{N}} : d ( \mathbf x , \mathbf 0 ) < \tfrac 12 \}$$ is open in the uniform topology, but not in the product topology.

    As $V$ is just the uniform $\frac{1}{2}$-ball centred at $\mathbf{0}$, it is clearly open in the uniform topology. It is not open in the product topology because there are no $n_1 , \ldots , n_k \in \mathbb{N}$ such that the set $\{ \mathbf{x} = ( x_n )_{n \in \mathbb{N}} : x_{n_1} = \cdots = x_{n_k} = 0 \}$ is a subset of $V$. (If it were open, then as $\mathbf{0} \in V$ there would be a basic open set $W = \prod_{n \in \mathbb{N}} W_n$, where each $W_n$ is open in $\mathbb{R}$ and $W_n = \mathbb{R}$ for all but finitely many $n$, such that $\mathbf{0} \in W \subseteq V$. If $n_1 , \ldots , n_k$ were those $n$ such that $W_n \neq 0$, then $\{ \mathbf{x} = ( x_n )_{n \in \mathbb{N}} : x_{n_1} = \cdots = x_{n_k} = 0 \} \subseteq W \subseteq V$.)

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The set $V$ isn’t the uniform $\frac12$-ball centred at $\mathbf{0}$ and in fact isn’t open in the uniform topology: see this answer. – Brian M. Scott Apr 23 at 10:24

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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