# Metric Spaces of Bounded Sequences

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?

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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.

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.

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