Let $\left\{ \ \left( X_\alpha, d_\alpha \right) \ \colon \ \alpha \in J \ \right\}$ be a collection of metric spaces, where $J$ is some (non-empty) set of indices. Let $$X \colon= \prod_{\alpha \in J} X_\alpha.$$ Let $\mathscr{T}_1$, $\mathscr{T}_2$, and $\mathscr{T}_3$ denote, respectively, the product, uniform, and box topologies on $X$. Here the uniform metric $\bar{\rho}$ on $X$ is defined as follows: Let $\mathbf{x} \colon= \left( x_\alpha \right)_{\alpha \in J}$ and $\mathbf{y} = \left( y_\alpha \right)_{\alpha \in J}$ be any elements of $X$. Then we define $$ \bar{\rho}( \mathbf{x}, \mathbf{y} ) \colon= \sup \left\{ \ \min \left\{ \ d_\alpha \left( x_\alpha, y_\alpha \right), \ 1 \ \right\} \ \colon \ \alpha \in J \ \right\}.$$ Can we show that $\mathscr{T}_1 \subset \mathscr{T}_2 \subset \mathscr{T}_3$? And if so, then can we show that these three topologies are all different (i.e. distinct) if the index set $J$ is infinite?
My effort:
Let $\mathbf{B} \colon= \prod_{\alpha \in J} B_\alpha$ be a basis element for the product topology on $X$, and suppose $ \mathbf{x} \colon= \left( x_\alpha \right)_{\alpha \in J} \in \mathbf{B}$. Let $\alpha = \alpha_1, \ldots, \alpha_n$ be the indices for which $B_\alpha \neq X_\alpha$; thus we can suppose that, for each $i = 1, \ldots, n$, the set $B_{\alpha_i}$ is an open set in $X_{\alpha_i}$; for all other indices $\alpha \in J$, we have $B_\alpha = X_\alpha$.
For each $i = 1, \ldots, n$, since $x_{\alpha_i} \in B_{\alpha_i}$ and since $B_{\alpha_i}$ is open in $X_{\alpha_i}$, therefore we can find a real number $\delta_i > 0$ such that the open ball $\mathrm{B}_{d_{\alpha_i}}\left( x_{\alpha_i}; \delta_i \right)$ is contained in $B_{\alpha_i}$, where $$ \mathrm{B}_{d_{\alpha_i}}\left( x_{\alpha_i}; \delta_i \right) \colon= \left\{ \ v_{\alpha_i} \in X_{\alpha_i} \ \colon \ d_{\alpha_i} \left( v_{\alpha_i}, x_{\alpha_i}\right) < \delta_i \ \right\} ;$$ we can also assume without any loss of generality that this $\delta_i$ satisfies $0 < \delta_i < 1$.
Let us now put $$\delta \colon= \min \left\{ \ \delta_1, \ldots, \delta_n \ \right\}. $$ Then $0 < \delta < 1$.
Now if $\mathbf{y} \colon= \left( y_\alpha \right)_{\alpha \in J} \in X$ and if $\bar{\rho}( \mathbf{x} , \mathbf{y} ) < \delta$, then, for each $\alpha \in J$, we have $$ \min \left\{ \ d_\alpha \left( x_\alpha, y_\alpha \right), \ 1 \ \right\} < \delta,$$ and hence $$ d_\alpha \left( x_\alpha, y_\alpha \right) < \delta,$$ because $0 < \delta < 1$. In particular, for each $i = 1, \ldots, n$, we have $$ d_{\alpha_i} \left( x_{\alpha_i}, y_{\alpha_i} \right) < \delta \leq \delta_i. $$
This shows that $y_{\alpha_i}$ is in the open ball $\mathrm{B}_{d_{\alpha_i}} \left( x_{\alpha_i}, \delta_i \right)$ and hence $y_{\alpha_i} \in B_{\alpha_i}$ for each $i = 1, \ldots, n$. So $y \in \mathbf{B}$.
Thus, for each basis element $\mathbf{B}$ for the product topology $\mathscr{T}_1$ on $X$, and for each element $\mathbf{x} \in \mathbf{B}$, we can find a basis element $\mathrm{B}_\bar{\rho} ( \mathbf{x} ; \delta)$, where $$ \mathrm{B}_\bar{\rho} ( \mathbf{x} ; \delta) \colon= \left\{ \ \mathbf{v} \in X \ \colon \ \bar{\rho} ( \mathbf{v}, \mathbf{x} ) < \delta \ \right\}, $$ for the uniform topology $\mathscr{T}_2$ containing $\mathbf{x}$ such that $$\mathrm{B}_\bar{\rho} ( \mathbf{x} ; \delta) \subset \mathbf{B}.$$ Therefore $$ \mathscr{T}_1 \subset \mathscr{T}_2. $$
Is what I've done so far all correct and accurate? If so, then is my presentation clear and my formatting consistent enough?
Now let $\varepsilon$ be a positive real number, let $\mathbf{x} \in X$, and let $\mathbf{y} \in \mathrm{B}_\bar{\rho} ( \mathbf{x} ; \varepsilon)$. Then $$ \bar{\rho} \left( \mathbf{x}, \mathbf{y} \right) < \varepsilon, $$ and so we can choose a real number $\delta$ such that $$ 0 < \delta < \min \left\{ \varepsilon - \bar{\rho} \left( \mathbf{x}, \mathbf{y} \right), \ 1 \ \right\}. $$ This number $\delta$ is such that $0 < \delta < 1$ and such that $$\mathrm{B}_\bar{\rho} ( \mathbf{y}; \delta) \subset \mathrm{B}_\bar{\rho} ( \mathbf{x} ; \varepsilon). $$
Now let $$ \mathbf{B}^\prime \colon= \prod_{\alpha \in J} \mathrm{B}_{d_\alpha}\left( y_\alpha; \frac{\delta}{2} \right),$$ where, for each $\alpha \in J$, the symbol $\mathrm{B}_{d_\alpha}\left( y_\alpha; \frac{\delta}{2} \right)$ denotes the open ball in the metric space $\left( X_\alpha, d_\alpha \right)$ of radius $\frac{\delta}{2}$ and centered at $y_\alpha$.
If $ \mathbf{z} \colon= \left( z_\alpha \right)_{\alpha \in J} \in \mathrm{B}^\prime$, then $\mathbf{z} \in X$ and, for each $\alpha \in J$, we have $$ d_\alpha \left( z_\alpha, y_\alpha \right) < \frac{\delta}{2} < \delta < 1,$$ and thus we also have $$ \min \left\{ \ d_\alpha \left( z_\alpha, y_\alpha \right) , \ 1 \ \right\} < \frac{\delta}{2}, $$ and so $$ d_\alpha \left( z_\alpha, y_\alpha \right) = \min \left\{ \ d_\alpha \left( z_\alpha, y_\alpha \right) , \ 1 \ \right\}, $$ which implies that $$ \bar{\rho}( \mathbf{z}, \mathbf{y} ) = \sup \left\{ \ \min \left\{ \ d_\alpha \left( z_\alpha, y_\alpha \right) , \ 1 \ \right\} \ \colon \ \alpha \in J \ \right\} \leq \frac{\delta}{2} < \delta, $$ and theefore $$ \mathbf{z} \in \mathrm{B}_\bar{\rho} \left( \mathbf{y}, \delta \right), $$ and hence $$ \mathbf{z} \in \mathrm{B}_\bar{\rho} \left( \mathbf{x}, \varepsilon \right), $$ thus showing that $$ \mathbf{B}^\prime \subset \mathrm{B}_\bar{\rho} \left( \mathbf{x}, \varepsilon \right). $$ Moreover as $$ d_\alpha \left( y_\alpha, y_\alpha \right) = 0 < \frac{\delta}{2} $$ holds for each $\alpha \in J$, so we can conclude that $\mathbf{y} \in \mathrm{B}^\prime$ also.
Thus we have shown that for each basis element $\mathrm{B}_\bar{\rho} \left( \mathbf{x}; \varepsilon \right)$ for the uniform metric topology $\mathscr{T}_2$ on $X$, and, for each point $ \mathbf{y} \in \mathrm{B}_\bar{\rho} \left( \mathbf{x}; \varepsilon \right)$, we can find a basis element $\mathrm{B}^\prime$ for the box topology $\mathscr{T}_3$ on $X$ such that $$ \mathbf{y} \in \mathrm{B}^\prime \subset \mathrm{B}_\bar{\rho} \left( \mathbf{x}; \varepsilon \right).$$ So $\mathscr{T}_2 \subset \mathscr{T}_3$.
Is what I have done so far totally right? If so, then have I managed to present my proof well enough? Is my formatting consistent enough?
Or, are there any problems with either of my proofs?
Is the above reasoning correct? Or, is there anything superfluous in it?