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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, [itex]L[/itex], of all bounded sequences with the metric [itex]\rho_1[/itex]

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

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

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

I believe convergence means:

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

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

Could anyone help me out?

Consider the space, [itex]L[/itex], of all bounded sequences with the metric [itex]\rho_1[/itex]

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

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

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

I believe convergence means:

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

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

Could anyone help me out?

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