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I've come across the following claim:

Let $\mathbb{R}^\infty = \cup_{n \geq 1} \mathbb{R}^n$ with the weak topology. Then

$((0, \dots, a_n, \dots, 0))_i$ converges implies $a_n = 0$ for all $n$ large.

I've been thinking about why there are two indexes, $i$ and $n$. My question is:

does it make sense to have 2 indexes? Or should it be $$\lim_{i \rightarrow \infty} ((0, \dots, a_i, \dots, 0))_i $$ instead of $$\lim_{i \rightarrow \infty} ((0, \dots, a_n, \dots, 0))_i $$?

My thought have led me to the conclusion that both make sense and that the claim is true for both. Am I right?

Many thanks for your help!

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the $i$ just indicates that it is a sequence where $a_n$ is a element in every touple of your sequence you are looking at. –  Listing May 7 '11 at 10:30
    
"Then $((0,\ldots,a_n,\ldots,0))_i$ converges implies $a_n=0$ for all $n$ large." makes no sense. What is $n$ large? What does $(0,\ldots,a_n,\ldots,0)$ mean? The it begins as zero and becomes eventually zero? Does it change anywhere at all? –  Asaf Karagila May 7 '11 at 13:53

2 Answers 2

up vote 8 down vote accepted

'Weak topology' has more than one meaning; is your $\mathbb R^\infty$ the space of sequences in $\mathbb R^\omega$ that are eventually $0$, topologized so that $U \subseteq \mathbb R^\infty$ is open iff each $U \cap \mathbb R^n$ is open in $\mathbb R^n$, where $\mathbb R^n$ is identified with $\{(x_i:i \in \omega) : \forall i \ge n (x_i = 0)\}$? If so, I can interpret $(0,\dots, a_n,\dots,0)$ as a sloppy notation for a sequence $(a_0,a_1,\dots)$ in which at most the term $a_n \ne 0$. I still don't see any reasonable interpretation of the original doubly-indexed version, but the singly-indexed version can be understood in several ways. Perhaps the most straightforward is as follows.

For $n \in \omega$ let $\alpha_n = (a^n_i:i \in \omega) \in \mathbb R^\infty$. Suppose further that for each $n \in \omega$ there is an index $\iota(n) \in \omega$ such that $a^n_i = 0$ whenever $i \neq \iota(n)$. The claim would then be that if $(\alpha_n:n \in \omega)$ converges, then $a^n_{\iota(n)} = 0$ for all sufficiently large $n$. On this interpretation the claim is false: just take $\alpha_n = (1,0,0,\dots)$ for all $n$.

In the special case in which the range of the function $\iota$ is infinite, however, the result is correct. Suppose that $(\alpha_n:n \in \omega)$ converges; clearly the limit must be the zero sequence $\zeta = (0,0,\dots)$. If the $a^n_{\iota(n)}$ are not eventually $0$, there is an infinite $I \subseteq \omega$ such that $\iota \restriction I$ is 1-1 and $\forall n \in I (a^n_{\iota(n)} \ne 0)$. Let $U = \{(x_n:n \in \omega) \in \mathbb R^\infty: \forall n \in I(|x_n| < |a^n_{\iota(n)}|)\}$; $U$ is an open nbhd of $\zeta$, but $\forall n \in I(\alpha_n \notin U)$, $\rightarrow\leftarrow$.

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Your notation is a little sloppy in the first place, writing $\mathbb R^{\infty} = \bigcup_{n \ge 1} \mathbb R^n$ means that an element of $\mathbb R^{\infty}$, as defined in this manner, must be in one of the $\mathbb R^n$ for some $n$ ; it doesn't make sense to speak of convergence in such a space, which to be honest must not have a very nice topology, whatever the "weak topology" term you used meant in that sentence. (I know what a weak topology is, but I don't think what you wrote was what you actually meant... )

If you meant that you wanted to think as if $\mathbb R \subset \mathbb R^2 \subset \dots \subset \mathbb R^n \subset \dots$, then you should've written $\mathbb R^{\infty} = \lim_{n \to \infty} \mathbb R^n$, or maybe you meant that your space looks more like this : $$ \mathbb R^{\mathbb N} = \{ \{x_n\} : \mathbb N \to \mathbb R \, | \, n \mapsto x_n \} $$ which basically means that your space is the space of all real sequences... (writing $\mathbb R^{\infty}$ is an ambiguous notation since the size of the infinity, i.e. its cardinal, is not clear in the notation! (though it is in the definition you wrote)) but then I don't know any "standard" way to bring a topology upon this space. It would be easier to make it clear in your mind if you actually told us what you were reading when you came across this or what you want to understand by working out this notation.

Note that there are many ways to speak of convergence when you are in infinite dimension, since you can now define multiple norms on the same space, which you can't in finite dimension (by that, I mean that all norms are equivalent for convergence...). That is one point that you must clear out to speak of convergence ; which norm are you using?

Hope that helps,

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2  
Actually, $\lim_{n\to\infty}\mathbb{R}^n$ is not exactly equal to $\mathbb{R}^{\mathbb{N}}$; the former would be the set of all sequences that are eventually $0$; the latter the set of all sequences. –  Arturo Magidin Jun 13 '11 at 4:56
    
That's why I wrote "or" ; I didn't know what he actually meant by $\mathbb R^{\infty}$, so I just went out and said what I know. Although it's good you mentioned it for the sceptical readers. –  Patrick Da Silva Jun 13 '11 at 6:22
    
But you said "or use a standard notation for this space, which is...", which states that $\mathbb{R}^{\mathbb{N}}$ is "a standard notation" for the same space, $\lim_{n\to\infty}\mathbb{R}^n$. If you meant to give two alternative possibilities, your phrasing does not suggest that, it suggests you are giving two alternative notations for the same possibility. –  Arturo Magidin Jun 13 '11 at 16:10
1  
@Arturo : Thanks for the request of clarification. I think it's good now. –  Patrick Da Silva Jun 13 '11 at 20:02

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