Convergent Sequences on l infinity

Kreizhn
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Homework Statement


Define R^\infty_f = \{ (t^{(1}),t^{(2}), \ldots, ) |\; t^{(i}) \in \mathbb{R}\; \forall i, \; \exists k_0 \text{ such that } t^{(k})=0 \; \forall k\geq k_0 \}

Define l^\infty = \{ (t^{(1}),t^{(2}), \ldots, ) |\; t^{(i}) \in \mathbb{R}\; \forall i, \; \sup_{k\geq 1} | t^{(k})| < \infty \}

Observe that R^\infty_f is a linear subspace of l^\infty. Show that R^\infty_f is not closed in l^\infty, then show that the closure of R^\infty_f is the space c_0;

Homework Equations



The space c_0 is the set of all sequences converging to zero


The Attempt at a Solution



It's not too hard to show that this set is not closed. It suffices to show that there is a convergent sequence in l^\infty such that every term is in R^\infty_f, but whose limit is not in R^\infty_f. I constructed the following sequence

x_1 = (1, 0, \ldots, )
x_2 = (1, \frac{1}{2}, 0 , \ldots, )
\vdots
x_n = (1, \ldots, \frac{1}{n}, 0, \ldots} )

which converges to the point a = (1, \frac{1}{2}, \ldots, \frac{1}{n-1}, \frac{1}{n}, \frac{1}{n+1}, \ldots )

It's the closure part that I'm worried about. I'm not terribly sure how I would go about showing that...
 
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It will suffice to show that every sequence in c_0 is a limit of sequences in R_f. Do you agree?
 
Yes, since every point in c_0 will necessarily be the limit of some sequence in R_f. Though I think that this only shows that c_0 is a subset of the closure - not necessarily the whole closure.
 
Yes, but on the other hand, R_f clearly sits in c_0 (and c_0 is closed!).
 
True enough.

So I need to show that every sequence in c_0 is a limit of sequences in R_f.

How do I show that every sequence that converges to zero is the limit of a sequence. Indeed, what does it mean for a sequence to be a limit of another sequence?
 
Think of this as a problem set in an abstract normed space. What does it mean for a sequence {x_n} to converge to x?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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