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Let $\ell^{\infty}$ the metric space of bounded sequences of real numbers $(x)=\{x_1, x_2,...\}$ with the metric $$d_{\infty}(x, y)=\sup_{n\in\mathbb{N}}|x_i-y_i|$$

Let $$A=\{x\in \ell^{\infty}: \exists\,\, k\in\mathbb{N}\,\,\, \text{so that}\,\,\, x_n=0, \forall n\geq k\}$$

Need to prove that this set is not closed, ie the need to set a sequence that converges to the limit but not in the group A. I thank you to help me because I could not find such a sequence.

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2 Answers 2

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Let $e_k$ be the sequence whose $k$-th term is $1$, the others are $0$, and define $x_n:=\sum_{j=1}^nj^{-1}e_j$. Then $x_n\in A$, and converges in $\ell^\infty$ to the sequence $\sum_{j=1}^{+\infty}j^{-1}e_j$, which is not in $A$.

One can check that the closure of $A$ in $\ell^\infty$ is the subspace of sequences which converge to $0$.

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  • $\begingroup$ Davide f I'm not mistaken $$\sum_{j = 1}^{\infty} j^{-1} e_j=\{1/n\}_{n\in\mathbb{N}} $$ right? $\endgroup$ Apr 19, 2013 at 16:58
  • $\begingroup$ @RoinerSeguraCubero Right. $\endgroup$ Apr 19, 2013 at 17:00
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An easier way is to show that the complement is not open. Let A be the point in the complement, namely the point x whose n-th coordinate is $1/n$, then any open set containing x clearly contains a point of A (since no matter how small a neighborhood we make, we will always find a coordinate of x, where x is smaller ever after). So the complement of A is not open, and so the set A is not closed.

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