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Metric Spaces of Bounded Sequences

  • Thread starter octane90
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  • #1
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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?
 
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Answers and Replies

  • #2
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Where, [itex]\rho_\infty(x,y)=sup(x_t,y_t)[/itex].
I doubt that this is correct. It should be [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]
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 [itex]X_k=(x_k^n)_n[/itex], then we can say that [itex]X_k\rightarrow Y[/itex] if for each [itex]\varepsilon>0[/itex], there exists an N such that for all [itex]k\geq N[/itex] holds that [itex]\rho_1(X_k,Y)<\varepsilon[/itex]. That last inequality is of course the same as

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

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 $\rho_1$ which should be allowed.
This is of course incorrect, since [itex]x_n=\sin(n)[/itex] is only one sequence. Again: you must have a sequence of sequences.
 
  • #3
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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.
 
  • #4
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The counterexample you're looking for is very easy. Think about sequences with only 0's and 1's.
 

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