Is ${\cal B}=\{ (p_1,q_1)\times (p_2,q_2)\times ...| \, p_i-q_i=p_j-q_j \}$ a basis for product topology? There is a problem from my topology course. For a collection ${\cal B}=\{ (p_1,q_1)\times (p_2,q_2)\times ...| \, p_i-q_i=p_j-q_j \, \forall i,j \in \mathbb{N}\}$.
Prove or Disprove: ${\cal B}$ is a basis for a topology on $\mathbb{R}^\omega$.
I know that for a product topology, the basis is ${\cal B}=\{(a_1,b_1)\times...\times(a_n,b_n)\subset\mathbb{R}^{\omega}|a_i<b_i\}$, but I'm not quite sure if the claim is true. I think it is not true and have disproved it in some way, but I don't think it seems quite right. Could you give me a hand?
 A: You have to check that two conditions are met:


*

*$\cup \mathcal{B} = X$.

*$\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2: \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$.


If these hold for a family $\mathcal{B}$ of subsets of $X$, then this family forms a base for some topology on $X$. And these conditions are both necessary and sufficient.
Clearly for your family the first condition is met: for every $x = (x_1, x_2, \ldots) \in \mathbb{R}^\omega$ we can just take $(x_1 - 1, x_1 + 1) \times (x_2 - 1, x_2 + 1) \times \ldots \in \mathcal{B}$ which contains $x$.
As to the second, consider the sets $B_1 = \prod_{n=1}^\infty (-\frac{1}{n+1}, 1-\frac{1}{n+1})$ (indeed in $\mathcal{B}$ as all intervals have length 1) and $B_2 = \prod_{n=1}^\infty (-1+\frac{1}{n+1}, \frac{1}{n+1})$, which is in $\mathcal{B}$ for the same reason. Both contain $0 = (0,0,0,\ldots)$. But $B_1 \cap B_2 = \prod_{n=1}^\infty (-\frac{1}{n+1}, \frac{1}{n+1})$ and there is no member of $\mathcal{B}$ that contains $0$ and is a subset of this intersection (why?). 
So it's indeed not a base for a topology, as you suspected.  
