# Closure of $\{ (x_n) | x_n = 0$ for almost all $n \}$

What is the closure of $\{ (x_n) | x_n = 0$ for almost all $n \}$ in $\mathbb{R}^\mathbb{N}$ with the product topology?

I'm not sure how to think about it. A point $x$ is in the closure of a set $A$ if for every neighbourhood $N$ of $x$, $N \cap A \neq \emptyset$.

But I'm not sure what a neighbourhood of a point in $\mathbb{R}^\mathbb{N}$ looks like. In fact, I'm not even sure what an open ball looks like.

Can someone tell me what open balls or neighbourhoods look like?

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Do you know what the definition of the product topology is? –  Qiaochu Yuan Feb 1 '11 at 11:39
The product topology is the topology generated by the basis that consists of all products such that only finitely many components are not the entire space. Those finitely many components that are not the entire space are open sets in the topology on the respective space of that component. –  Rudy the Reindeer Feb 2 '11 at 11:56

A better way to think about $\mathbb{R^N}$ is as the set of all real-valued sequences, right? You have $\mathbb N$ dimensions and each dimension corresponds to a real number, so $\mathbb{R^N}$ is just the set of all real-valued sequences (as a note $\mathbb{R^R}$ can be thought of as all the functions from $\mathbb R$ to $\mathbb R$).

So with this definition in mind, what does it mean for there only be a finite number of non-zero indexes in a sequence. Well it means that past a certain point the sequence is entirely zeros. Okay now keeping this in mind, think about what the product topology means. Well it means that any open set (We'll just work with basis sets) is a product of spaces $X_i$ where for only finitely many $i$ $X_i\neq \mathbb R$. Then if we want to show that our set is dense in $\mathbb{R^N}$, we need to show that it intersects every basis open set. Well we've just said that an open set has only finite many indices that aren't the whole space. So if we pick an $x \in X_i$ where each $X_i \neq \mathbb R$, then we'll get finitely many points that aren't zero and we can let everything else be zero. This will show that our set is dense in $\mathbb{R^N}$.

What happens if we put the box topology on $\mathbb{R^N}$?

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In the box topology it doesn't work anymore because I can find a basis set that consist of open intervals, none of which contains 0 in any of its components. Therefore not every basis set contains a point in $A$. –  Rudy the Reindeer Feb 2 '11 at 12:16
That is correct, it is not dense in the box topology. –  JSchlather Feb 2 '11 at 22:34

@Matt: Recall that in the product topology there are only finitely many restrictions (i.e finite number of open sets). The answer is that it is dense in the product topology. Can you see why? Pick a basic open set in the product topology and try to find a point which is in your set and in the basic open set. It helps to think about 0's.

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@Matt: It is closed in the box topology. Try to show the complement is open, it helps to think about the open set $\mathbb{R} \setminus \{0\}$ when the coordinate is non-zero. –  user6495 Feb 2 '11 at 16:16