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Hi everyone this is my first post so apologies in advance if I do anything wrong...

I'm working through an unmarked assignment sheet and am struggling on this question:

Let $X = \prod^{\infty}_{i=1} X_i$ where $X_i = \mathbb{R}$ for all $i$.

Let $A=\{x=(x_i) \in X$: there exists $N \geq 1$ such that $x_i =0$ for all $i\geq N \}$.

Compute the closure of A in the product topology and the box topology. (Hint: A is closed in one of the topologies. In the other topology, every nonempty basis element intersects A.)

My thinking is that the closure of A in the product topology is $X$ and in the box topology it is A. However I am uncertain on the validity of how I got to this, if it is even right.

For the product topology I have a basis, (with $(a,b)_i $ the open interval in $X_i$)

$ B=\{\prod^\infty_{i=1}(a,b)_i : (a,b)_i = (-\infty , \infty) \text{ for all but finitely many } i \} $.

I think this satisfies the hint that one of the topologies has the property that every nonempty basis element intersects A but then with $(a,b)_i \neq (-\infty , \infty)_i \forall i$ the product does not necessarily intersect with A?

For the box topology I'm not sure what the best approach is, I just cheated and assumed the other topology was closed and thus =A.

I would really appreciate some help understanding this. Thanks.

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Very good first post. –  Brian M. Scott Feb 24 '13 at 16:40
    
@BrianM.Scott Thank you, I appreciate everyones help. –  120 Eyes Feb 24 '13 at 16:56

1 Answer 1

up vote 5 down vote accepted

The box topology is finer (has more open sets) than the product topology. So if $A$ were closed in the product topology its complement would be open in that topology and it would remain open in the box topology, and so $A$ would still be closed in the box topology. So the meta-hint gives away that $A$ must be closed in the box topology, and dense in the product topology...

Suppose we have a point $(x_n)$ not in $A$. This means that there is no tail of the coordinates that is $0$ and this means, if you think about it, exactly that there are infinitely many coordinates $k$ such that $x_k \neq 0$. We can find open sets around those coordinates that also miss $0$ and use those to build a box-open basic set that has the property that all its members are also non-zero at these same coordinates.

When we have a basic product open set, we only have "control" over finitely many coordinates, and all values beyond those finitely can be anything (all of $\mathbb{R}$ is allowed). In particular, they could all be $0$...

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That has cleared up a lot of things, thanks for your help. –  120 Eyes Feb 24 '13 at 15:42
    
@120Eyes You're welcome! –  Henno Brandsma Feb 24 '13 at 15:44

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