Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

2 Answers 2

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.

share|improve this answer

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.