Is the set $E$ of sequences containing only entries $0$ and $1$ in $(m,\left \| \cdot \right \|_\infty)$ complete?

I can't really wrap my head around $E$, or a Cauchy sequence in $E$. I need to take a Cauchy sequence in $E$ and show it's Cauchy in $(m,\left \| \cdot \right \|_\infty)$? I think I can show $(m,\left \| \cdot \right \|_\infty)$ is complete but I don't know how to use that info here.

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Can you show that it is closed in $m$? Then use completeness of $m$. [for those not acquainted with the notation, $m$ is more commonly denoted by $\ell^\infty$] –  t.b. Feb 16 '12 at 16:51

The functions $f_n : \ell^\infty \rightarrow \mathbb{R}, u \mapsto u_n$ are continuous (they're Lipschitz continuous), and the set $\{0,1\}$ is closed in $\mathbb{R}$, therefore $f_n^{-1}\{0,1\}$ is closed in $\ell^\infty$. Your set $E$ is the intersection $E = \bigcap_{n\in\mathbb{N}} f_n^{-1}\{0,1\}$ and is therefore closed in $\ell^\infty$. Since $\ell^\infty$ is a Banach space, and a closed subspace of a complete metric space is complete, $E$ is complete.

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Hint: Note that if a sequence is Cauchy in $E$, then it is Cauchy in $\ell_\infty$, and thus converges in $\ell_\infty$.

If a sequence $(c^j)_{j=1}^\infty$in $E$ converged to an element $x$ in $\ell_\infty\setminus E$, take a non-zero coordinate $x_i$ of $x$ that is not equal to 1. Then consider the quantities $|x_i-c^j_i|$, $j=1,2,\ldots$.

More simply, you might observe that if a sequence is Cauchy in $E$, then it is eventually constant (the distance between two elements $c^1\ne c^2$ in $E$ is 1).

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