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Theorem: Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathcal{B}_{\alpha}$. The collection of all sets of the form $$\prod_{\alpha \in J} B_{\alpha}$$ where $B_{\alpha} \in \mathcal{B}_{\alpha}$ for each ${\alpha}$, forms a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$

My Attempted Proof

The topological space we are dealing with here is $\left(\prod_{\alpha \in J} X_{\alpha}, \mathcal{T}_B\right)$, where $\mathcal{T}_B$ is the box topology on $\prod_{\alpha \in J} X_{\alpha}$. Let $\mathcal{B}_{\text{box}}$ denote the standard basis for the box topology.

Put $$\mathcal{F} = \left\{ \ \prod_{\alpha \in H} B_{\alpha} \ \middle| \ B_{\alpha} \in \mathcal{B}_{\alpha} \ \text{for each $\alpha$} \ \right\}.$$ Then take $U \in \mathcal{T}_B$, and take $x \in U$. Since $U = \bigcup_{ B \in \mathcal{K}}B$ where $B = \prod_{\alpha \in J}U_{\alpha}$ and $U_{\alpha}$ is open in $X_{\alpha}$ and $\mathcal{K} \subset \mathcal{B}_{\text{box}}$ we then have $x \in B$ for some $B \in \mathcal{K}$. In other words $x \in \prod_{\alpha \in J}U_{\alpha}$ for some indexing set $J$.

Now since each $U_{\alpha}$ is open in $X_{\alpha}$, we have $U_{\alpha} = \bigcup_{\gamma \in I}B_{\gamma}$ where $B_{\gamma}$ are basis elements of $X_{\alpha}$. Therefore $$x \in \prod_{\alpha \in J}\left(\bigcup_{\gamma \in I}B_{\gamma}\right)_{\alpha}$$

By elementary set theory we have $$\left(\bigcup_{i \in I} V_i\right) \times \left(\bigcup_{j \in J} W_j\right) = \bigcup_{\langle i, j \rangle I \times J}\left(V_i \times W_j\right)$$

So that $$\prod_{\alpha \in J}\left(\bigcup_{\gamma}B_{\gamma}\right)_{\alpha} \ = \ \bigcup_{\langle i_1, i_2, ... \rangle \in I_1 \times I_2 \times ...} \underbrace{\left(B_{i_1} \times B_{i_2} \times ...\right)}_{|J| \ \text{times}}$$

Choose $H = I_1 \times I_2 \times I_3 \times ...$

Then $\exists \prod_{\alpha \in H}B_{\alpha} \in \mathcal{F}$ such that $x \in \prod_{\alpha \in H}B_{\alpha}$. It also follows that $$\prod_{\alpha \in J}\left(\bigcup_{\gamma}B_{\gamma}\right)_{\alpha} \subset U$$ Since $\prod_{\alpha \in J}\left(\bigcup_{\gamma}B_{\gamma}\right)_{\alpha} = \prod_{\alpha \in J}U_{\alpha} = B \subset U$. Thus it follows that $\mathcal{F}$ is a basis for the box topology on $\prod_{\alpha \in J} X_{\alpha}$. $\square$

Is my above proof correct and rigorous? Are there areas of it that I can improve on? Can this be proved in an easier way? I apologize in advance if the proof is long or messy (especially with the indices)

If my proof is incorrect however, or if it is not rigorous please let me know!

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    It looks good to me! Just how bogged down you want to get in the set-theory statements can be a matter of taste, I suppose. – 2017-01-18
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    I don't why you have to prove it. It's the def'n for the "canonical" base for the box-product topology. – 2017-01-18

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Suppose $x =(x_i)_i \in O$, where $O$ is box open. This means that there are open sets $O_i \subseteq X_i$, such that $x \in \prod_i O_i \subseteq O$, by the definition of the box topology (topology generated by open "boxes", so the latter form a base). As each $\mathcal{B}_i$ is a base in $X_i$, we pick $B_i \in \mathcal{B}_i$ such that $x_i \in B_i \subseteq O_i$, but then $x \in \prod_i B_i (\in \mathcal{F}) \subseteq \prod_i O_i \subseteq O$. So sets from $\mathcal{F}$ form a base.

There is no need for set-theory manipulation here, just standard definitions.