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?

• You don't need to (explicitly) think of the product topology at all. You just need to show that $\mathcal{B}$ is (or is not) a basis for $some$ topology. There is a simple criterion to show that something is a basis for a topology (that I imagine should be in your notes somewhere), try looking for it. – James Dec 17 '14 at 12:28
• Is it "for every open set $U$ and every $x\in U$, there exist a element $C$ of ${\cal C}$ such that $x\in C\subset U$, then ${\cal C}$ is a basis"? – Eric Curtiss Dec 17 '14 at 12:34
• Not quite that, as, you need to know what topology you are a base for to check that condition. What I had in mind is that the elements of $\mathcal{B}$ cover $\mathbb{R}^\omega$, and for each $B_1,B_2 \in \mathcal{B}$, and each $x \in B_1 \cap B_2$, there is $B_3 \in \mathcal{B}$, so that $x \in B_3 \subseteq B_1 \cap B_2$. You can see the wikipedia page for Base (topology) for exposition. – James Dec 17 '14 at 12:45

You have to check that two conditions are met:

1. $\cup \mathcal{B} = X$.
2. $\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2: \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$.

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.

• Because the length of each intervals $(-\frac{1}{n+1}, \frac{1}{n+1})$ is going to infinitely small, so an actual length of intervals in that subset can not exist? – Eric Curtiss Dec 19 '14 at 7:21
• Yes, if such a $B_3$ would exist, then it is of the form $\prod_{n=1}^\infty (p_n,q_n)$ with $r = p_1 - q_1 = p_2 - q_2 = \ldots$ and $r>0$ so for some $n$ we have $\frac{1}{n+1} < r$, and so the $n$-th factor of $B_3$ cannot be a subset of the $n$-th factor of $B_1 \cap B_2$ from that coordinate onwards. – Henno Brandsma Dec 19 '14 at 15:46