2
$\begingroup$

I want to prove the following theorem:

If the topology on each space $X_\alpha$ is generated by a basis $\mathcal{B}_{\alpha}$. The set $\mathcal{B}={\underset{\alpha\in J}{\prod}B_{\alpha}:\forall\alpha\in J,\,B_{\alpha}\in\mathcal{B_{\alpha}}} $ is therefore a basis for the box topology on $\underset{\alpha\in J}{\prod}X_{\alpha} $.

My attempt: Let $U$ be an open set in the box topology of $\underset{\alpha\in J}{\prod}X_{\alpha} $. I want to prove that it can be written as the union of elements of $\mathcal{B} $. We can write $U$ as the union of elements of the standard basis for the box topology: $U=\underset{i\in I}{\bigcup}\underset{\alpha\in J}{\prod}U_{\alpha}^{i} $ (with each $U_{\alpha}^{i} $ an open set of $X_\alpha$). Now, each $U_{\alpha}^{i} $ can be written as the union of elements of $\mathcal{B}_{\alpha} $: $U_{\alpha}^{i}=\underset{\lambda\in\Lambda}{\bigcup}B_{\alpha,\lambda}^{i} $. Thus, we get $U=\underset{i\in I}{\bigcup}\underset{\alpha\in J}{\prod}\underset{\lambda\in\Lambda}{\bigcup}B_{\alpha,\lambda}^{i}=\underset{i\in I}{\bigcup}\underset{\lambda\in\Lambda}{\bigcup}(\underset{\alpha\in J}{\prod}B_{\alpha,\lambda}^{i}) $. The factor in parentheses is an element of $\mathcal{B} $.

I don't know how to get $U$ as an union of such elements of $\mathcal{B}$ in order to conclude the proof.

2 Answers 2

1

As concerns your attempt, the family $\Lambda$ such that $U_\alpha^i=\bigcup_{\lambda\in\Lambda}B^i_{\alpha,\lambda}$ depends on $\alpha$, so it is appropriate to denote it $\Lambda_\alpha$. Then you can see that it has no sense to write $\bigcup_{\lambda\in\Lambda_{\alpha}}\big(\prod_{\alpha\in J}B^i_{\alpha,\lambda}\big)$, as you did. So, your last equality is not valid.

The term $\prod_{\alpha\in J}\bigcup_{\lambda\in\Lambda_\alpha} B^i_{\alpha,\lambda}$ denotes the family of all functions $f$ with domain $J$ such that $f(\alpha)$ is an element of $\bigcup_{\lambda\in\Lambda_\alpha}B^i_{\alpha,\lambda}$, for each $\alpha\in J$. For any such $f$ we can find a function $g$ mapping each $\alpha\in J$ to $g(\alpha)\in\Lambda_\alpha$ such that $f(\alpha)\in B^i_{\alpha,g(\alpha)}$, thus $g\in\prod_{\alpha\in J}\Lambda_\alpha$ and $f\in\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)}$. Also, for any $g\in\prod_{\alpha\in J}\Lambda_\alpha$, each $f\in\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)}$ is an element of $\prod_{\alpha\in J}\bigcup_{\lambda\in\Lambda_\alpha} B^i_{\alpha,\lambda}$. So we can write $$\prod_{\alpha\in J}\bigcup_{\lambda\in\Lambda_\alpha} B^i_{\alpha,\lambda}= \bigcup_{g\in\prod_{\alpha\in J}\Lambda_\alpha}\quad\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)},$$ which is a union of elements of $\mathcal{B}$.

  • 0
    Don't we get $$U=\bigcup_{i\in I}\bigcup_{g\in\prod_{\alpha\in J}\Lambda_\alpha}\quad\prod_{\alpha\in J} B^i_{\alpha,g(\alpha)}$$ in that case? It is still not an union of elements of $\mathcal{B}$... – 2017-01-17
  • 0
    $\prod_{\alpha\in J}B^i_{\alpha,g(\alpha)}$ is an element of $\mathcal{B}$, so $U$ is a union of elements of $\mathcal{B}$. – 2017-01-17
  • 0
    Right: the union of the union is still an union, of course. – 2017-01-17
  • 0
    Shouldn't $\Lambda$ be denoted $\Lambda_{\alpha}^{i}$? – 2017-01-17
  • 0
    Yes, you are right. – 2017-01-17
2

A common method for proving that a set $U$ is a union of elements of a collection $\mathcal B$ is to prove that for each point $x \in U$ there exists $B \in \mathcal B$ such that $x \in B \subset U$.

So suppose that $x \in U$. In more detail, $x=(x_\alpha)_{\alpha \in J} \in U = \bigcup_{i \in I} \prod_{\alpha \in J} U^i_\alpha$.

Then there exists $i \in I$ such that $x \in \prod_{\alpha \in J} U^i_\alpha$.

It follows that for each $\alpha \in J$ we have $x_\alpha \in U^i_\alpha$.

Because $\mathcal B_\alpha$ is a basis for $X_\alpha$, it follows that there exists $B^i_\alpha \in \mathcal{B}_\alpha$ such that $x_\alpha \in B^i_\alpha \subset U^i_\alpha$.

Therefore $x \in B^i \equiv \prod_{\alpha \in J} B^i_\alpha \subset \prod_{\alpha \in J} U^i_\alpha \subset U$, and $B^i \in \mathcal{B}$.