Product space and Basis

Question

If for each i , $\mathbb{B}_i$ is a basis for the topology of $X_i$, then the set $\{ B_1\times.....\times B_n:B_i \in \mathbb{B}_i\}$ is a basis for the product topology on $X_1\times...\times X_n$

Show that the above holds for $n=2$.

Attempt:

Let $n=2$. $\{B_1 \times B_2:B_i \in \mathbb{B}_i\}$ is clearly open in $X_1\times X_2$. Let $W$ be open in the product topology, so $W=\bigcup_{i \in I}(U_i \times V_i)$ for some index set $I$. Since $U_i$ is open in $X_1$, $U_i=\bigcup_{k \in K_i}B_k$ (open in $X_1$) and for a similar reason, $V_i=\bigcup_{m \in M_i}B_{M}$ (open in $X_2$) Hence $W= \bigcup_{i \in I,k\in K_i,m\in M_i}(B_{k}\times B_m)$

Is this proof correct? (Please answer this question then add feedback, please)

Answer 1

It's mostly correct, but since you're asking for suggestions, I'll be picky.

You write that $$ \{ _1 \times _2 : _ \in _ \}$$ is clearly open in $_1\times _2.$

That's not true. Each of the individual sets $B_1 \times B_2$ is clearly open, but the collection of all of those things ... that's not even a subset of $X_1 \times X_2$, so it cannot be open.

The second problem is that you're overusing the symbol $B_i$. Suppose for a moment that $\Bbb B_1$ has three elements, and $\Bbb B_2$ has two. You've written, for instance, that

$$ U_1 = \cup_{k \in K_1} B_k. $$ So supposing that $K_1$ contains the numbers $2$ and $3$, you have $$ U_1 = B_2 \cup B_3 $$ and have noted that $B_2$ and $B_3$ are open in $X_1$. Then for $V_1$, you've written something similar, for which perhaps we end up with $$ V_1 = B_1 \cup B_2 $$ where $B_1$ and $B_2$ are open in $X_2$. But $B_2$ is already an open set in $X_1$. So which is it? When you get right down to it, it's a mess.

Here's a rewrite:

We'd like to show that $Q = \{B_1 \times B_2:B_i \in \mathbb{B}_i\}$ is a basis for the product topology for $X_1 \times X_2$. Clearly each element of $Q$ is open in $X_1\times X_2$.

Suppose $W$ be open in the product topology. Then $W=\bigcup_{i \in I}(U_i \times V_i)$ for some index set $I$, where each $U_i$ is open in $X_1$ and each $V_i$ is open in $X_2$.

Since $U_i$ is open in $X_1$, there is an index set $K_i$ such that $$ U_i=\bigcup_{k \in K_i} C_k $$ where each $C_k$ is in $\Bbb B_1$; similarly, we can write $$ V_i=\bigcup_{m \in M_i}D_{m} $$ where each $D_m$ is open in $X_2$.

Substituting the expressions for $V_i$ and $U_i$, and collecting the union of unions into one large union, we get

$$ W = \bigcup_{i \in I, k\in K_i, m\in M_i}(C_{k}\times D_m) $$ and because $C_k \times D_m$ is an element of $Q$, we see that $Q$ is a basis for the topology on $X_1 \times X_2$.

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I can't say whether that proof's perfect, but it clearly uses the same ideas you were trying to use, and it (mostly) gets rid of the notational conflicts.

Answer 2

It is correct, if you add that each $U_i$ is an open subset of $X_1$ and that each $V_i$ is an open subset of $X_2$.

Answer 3

It's correct as John and José already stated.

I would personally use a different approach, using an alternative characterisation of bases:

Then $O \subseteq X_1 \times \ldots \times X_n$ be open and $(x_1,\ldots,x_n) \in O$. By definition of the product topology, there are open sets $O_i$ for $i=1,\ldots,n$ such that $(x_1, \ldots,x_n) \in O_1 \times \ldots \times O_n \subseteq O$.

So for each $i$, $x_i \in O_i$ and as $\Bbb B_i$ is a base for $X_i$, there is a set $B_i \in \Bbb B_i$ such that $x_i \subseteq B_i \subseteq O_i$.

Then $B_1 \times \ldots \times B_n$ is a set from our set we want to prove to be a base and clearly

$$(x_1, \ldots,x_n) \in B_1 \times \ldots \times B_n \subseteq O_1 \times \ldots \times O_n \subseteq O$$

so we indeed have a base, as $O$ and its element $(x_1, \ldots,x_n)$ were arbitrary.

On notation: consider using $\prod_{i=1}^n B_i$ instead of the $B_1 \times \ldots \times B_n$ notation, it will be easier once you switch to infinite/arbitrary products.