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I was trying to extend the monotone convergence theorem for sequences in $\mathbb{R}$ to sequences in $\mathbb{R}^n$. The monotone convergence theorem for sequences in $\mathbb{R}$ is:

If $X=(x_n)_{n\in \mathbb{N}}$ is a non-decreasing sequence of real numbers, then the sequence $X$ converges if and only if it is bounded in which case $\lim_{n\to\infty} x_n=\sup_{n\in\mathbb{N}} x_{n}$

I think that in the context of $\mathbb{R}^n$, "non-decreasing" could mean that the sequence of the norms of the elements is non-decreasing, but it seems to me that this is not sufficient to guarantee convergence. I think that the notion of "direction" is important in this case and that "direction" can be made precise if we consider the dot product. More precisely, we could place some assumptions on the convergence of $x_{n+1}\cdot x_{n}$ as $n\to\infty$ (where "$\cdot$" is the dot product). Does anyone have an idea as to where I can find the general theorem in the context of $\mathbb{R}^n$. Thanks in advance.

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Having non-decreasing norms is not sufficient even in one dimension: consider the sequence 1, -1, 1, -1, ... –  Nate Eldredge Jun 12 '11 at 18:54
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up vote 5 down vote accepted

Perhaps you'll wish to consider the natural partial order on $\mathbb{R}^n$: $\bf{x} \le \bf{y}$ if and only if $x_i \leq y_i$ $\forall i=1,\ldots,n$.

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Above ${\bf{x}} = (x_1,\ldots,x_n)$ and ${\bf{y}} = (y_1,\ldots,y_n)$. –  Shai Covo Jun 12 '11 at 8:07
    
you are right! this is simpler and extend directly the definition of non decreasing in the sense you use it to each coordinate of the vector. –  Ivan3.14 Jun 12 '11 at 8:27
    
Thanks, and notice that you can apply the usual convergence theorem to each of the components separately (assuming the sequence of vectors is bounded). –  Shai Covo Jun 12 '11 at 8:32
    
Suggests that any ordered set with a least upper bound property will also have the property that every non-decreasing and bounded sequence will converge? –  user9352 Jun 17 '11 at 13:25
    
@user9352: That might be a good point, but I have no special knowledge of the subject... –  Shai Covo Jun 19 '11 at 4:58
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