Metric Spaces of Bounded Sequences

[octane90]

I was attempting to find a counterexample to the problem below. I think I may have, but was ultimately left with more questions than answers.

Consider the space, L, of all bounded sequences with the metric \rho_1

\displaystyle \rho_1(x,y)=\sum\limits_{t=1}^{\infty}2^{-t}|x_t-y_t|

Show that a sequence that converges to X^* in (L,\rho_1) does not necessarily converge to X^* in (L,\rho_\infty)

Where, \rho_\infty(x,y)=sup_t|x_t-y_t|.

I believe convergence means:

X converges to Y if \forall \epsilon>0 \exists N s.t. \sum\limits_{t=N}^{\infty}2^{-t}|x_t-y_t|<\epsilon

I believe that x_n=sin (n) converges to the 0 sequence in \rho_1, but not in \rho_\infty. However, it seems like x_n=sin (n) could also be shown to converge to {1,1,1,1,...} under \rho_1 which shouldn't be allowed.

Could anyone help me out?

 

[micromass]

Where, \rho_\infty(x,y)=sup(x_t,y_t).

I doubt that this is correct. It should be \rho_\infty(x,y)=\sup_t |x_t-y_t|.

I believe convergence means:

X converges to Y if \forall \epsilon>0 \exists N s.t. \sum\limits_{t=N}^{\infty}2^{-t}|x_t-y_t|<\epsilon

This is not what convergence is. First of all you must have a sequence of elements in L. That is, you have to start from a sequence of sequences. Let X_k=(x_k^n)_n, then we can say that X_k\rightarrow Y if for each \varepsilon>0, there exists an N such that for all k\geq N holds that \rho_1(X_k,Y)<\varepsilon. That last inequality is of course the same as

\sum_{t=1}^{+\infty} 2^{-t}|x_k^t-y^t|<\varepsilon

I believe that x_n=sin (n) converges to the 0 sequence in \rho_1, but not in \rho_\infty. However, it seems like x_n=sin (n) could also be shown to converge to {1,1,1,1,...} under $\rho_1$ which should be allowed.

This is of course incorrect, since x_n=\sin(n) is only one sequence. Again: you must have a sequence of sequences.

 

[octane90]

Thanks, that was very helpful. It makes sense that I need to be thinking of a sequence of sequences. I was just having trouble wrapping my head around this L space.

And you are of course right about my typo in the problem set-up, I fixed it.

 

[micromass]

The counterexample you're looking for is very easy. Think about sequences with only 0's and 1's.