# Thread: Box and Product basis

1. ## Box and Product basis

Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathfrak{B}_{\alpha}$. The collection of all sets of the form $\prod_{\alpha\in J}B_{\alpha}$, where $B_{\alpha}\in\mathfrak{B}_{\alpha}$ for each $\alpha$, will serve as a basis for the box topology on $\prod_{\alpha\in J} X_{\alpha}$.
The collection of all sets of the same form where $B_{\alpha}\in\mathfrak{B}_{\alpha}$ for finitely many indices of alpha and $B_{\alpha}=X_{\alpha}$ for all the remaining indices, will serve as a basis for the product topology on $\prod_{\alpha\in J} X_{\alpha}$.

How am I supposed to prove this?

2. ## Re: Box and Product basis

Originally Posted by dwsmith
Suppose the topology on each space $X_{\alpha}$ is given by a basis $\mathfrak{B}_{\alpha}$. The collection of all sets of the form $\prod_{\alpha\in J}B_{\alpha}$, where $B_{\alpha}\in\mathfrak{B}_{\alpha}$ for each $\alpha$, will serve as a basis for the box topology on $\prod_{\alpha\in J} X_{\alpha}$.
The collection of all sets of the same form where $B_{\alpha}\in\mathfrak{B}_{\alpha}$ for finitely many indices of alpha and $B_{\alpha}=X_{\alpha}$ for all the remaining indices, will serve as a basis for the product topology on $\prod_{\alpha\in J} X_{\alpha}$.

How am I supposed to prove this?
Why don't you tell us? What are the definitions of the relevant terms?

3. ## Re: Box and Product basis

Originally Posted by Drexel28
Why don't you tell us? What are the definitions of the relevant terms?
There is nothing really to define.

4. ## Re: Box and Product basis

Originally Posted by dwsmith
There is nothing really to define.
Ok, so what do you think we should do? We want to show that every open set in the product space can be written as a union of things our 'product basis'. How can we do that?

5. ## Re: Box and Product basis

Originally Posted by Drexel28
Ok, so what do you think we should do? We want to show that every open set in the product space can be written as a union of things our 'product basis'. How can we do that?
We can set U as the union of all the basis since it exists in tau.

6. ## Re: Box and Product basis

Originally Posted by dwsmith
We can set U as the union of all the basis since it exists in tau.
I'm sorry, I don't get what you mean by that.

7. ## Re: Box and Product basis

Originally Posted by Drexel28
I'm sorry, I don't get what you mean by that.
$(X,\tau)$

The union of members of tau are in tau. Therefore, the union of the basis elements are in tau. We can call that union U.

8. ## Re: Box and Product basis

Originally Posted by dwsmith
$(X,\tau)$

The union of members of tau are in tau. Therefore, the union of the basis elements are in tau. We can call that union U.
What you need to show is that for every [tex]V\in\tau[tex] one can write $V$ as a union of the product of elements of $\mathfrak{B}_\alpha$. How do you propose we do that?

9. ## Re: Box and Product basis

Originally Posted by Drexel28
What you need to show is that for every [tex]V\in\tau[tex] one can write $V$ as a union of the product of elements of $\mathfrak{B}_\alpha$. How do you propose we do that?
I don't know.

10. ## Re: Box and Product basis

there are a couple of things to note, here:

first of all, if the index set J is finite, you get the same topology in either case.

for reasons that are often not made clear from the definition itself, the product topology is the largest (finest) topology that makes the projection functions continuous.

so what can go wrong in the box topology? suppose you have $\mathbb{R}^\omega$.

clearly, the set $U = \prod_{n \in \mathbb{N}}(-1/n,1/n)$ is open in the box topology. if $f:\mathbb{R} \rightarrow \mathbb{R}^\omega$ is the diagonal function: f(x) = (x,x,....,x), each component function is continuous (the component functions are just the identity on $\mathbb{R}$).

but f itself is NOT continuous. suppose it were. the origin is in U, so a pre-image of U must contain some interval containing 0 (since we are supposing the pre-image is open, if f is to be continuous). but any such interval is mapped to its countable cartesian product of itself, and so for SOME n, is mapped outside of U.

so the pre-image of U is {0}, a closed set, and f is not continuous, every though every projection (component) of f is.

with an infinite cartesian product, the box topology has "too many sets". limiting the product topology to just finite products of base elements, makes it just small enough so the projection functions are continuous. this is typical of topology, you want to make the domain set toplogy "just fine enough" and the target (image space) set topology "just coarse enough". with a product space, we are given the target topologies (presumably they are given topologies on the spaces that form the product), so the only topology we have any real freedom to define, is on the product space.

11. ## Re: Box and Product basis

Originally Posted by Deveno
there are a couple of things to note, here:

first of all, if the index set J is finite, you get the same topology in either case. I understand this

for reasons that are often not made clear from the definition itself, the product topology is the smallest (coarsest) topology that makes the projection functions continuous.

so what can go wrong in the box topology? suppose you have $\mathbb{R}^\omega$.

clearly, the set $U = \prod_{n \in \mathbb{N}}(-1/n,1/n)$ is open in the box topology.
Why is that open in the box topology? The paragraph below I didn't really get.

12. ## Re: Box and Product basis

is not each interval (-1/n,1/n) a basis element (open interval) in the standard topology for the real numbers? we're just taking a (countably) infinte product of such open intervals, one for each n. each one is a basis set. such products of basis sets ARE the basis elements of the box topology. basis elements are open (every set in a basis also belongs to the topology, the topology is formed from (arbitrary) unions of these sets).

a continuous function f:X-->Y (where X,Y are topological spaces) is one where the pre-image of an open set U in the topology for Y, $f^{-1}(U)$, is open in the topology for X.

in my example, X is the real numbers. an open set contains a neighborhood (think: basis element) containing each of its points (if a set S is open, and x is in S, then there is a neighborhood containing x entirely within S). now the point (0,0,...,0) of $\mathbb{R}^\omega$ is clearly contained in U (since 0 is contained in each interval (-1/n,1/n)).

so $f^{-1}(0,0,...,0) = 0$ is contained in $f^{-1}(U)$. if this pre-image is open (which is has to be if f is to be continuous), there must be SOME interval K containing 0 contained within $f^{-1}(U)$, say (-ε,ε).

but now we're in a fix: having decided upon our epsilon, we can see that for large enough n, say n > M = 1/ε, (-1/n,1/n) does not contain (-ε,ε).

but surely, since $K \in f^{-1}(U)$, $f(K) \in f(f^{-1}(U)) = U$. however f(K) = (-ε,ε) x (-ε,ε) x.....x (-ε,ε) (countably many times), and this can't possibly lie within U (once we get past the M-th copy...we're in trouble).