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We define the following infinity norm on $\mathbb{R}$ as follows $$l_\infty(\mathbb{R}) = \{ (x_i)_{i \in \mathbb{N}} \,\mid\, x_i \in \mathbb{R}, \sup_{i\in\mathbb{N}} \left|x_i\right|<\infty \}$$

Now let $M \subset l_\infty(\mathbb{R})$ be the subspace consisting of all sequences $x =(x_i)_{i\in\mathbb{N}}$ with finitely many nonzero terms.

Now the questions asks me to Find a Cauchy sequence in M that does not converge in M.

I do not know how to approach this problem. Otherwise I would have tried, I have problems thinking about how the norm is defined on R, and all the symbols confuse me. Could someone help me break this down, into something very elementary and basic? =)

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What you quoted is not a norm, but the definition of $l_{\infty}(\mathbb R)$, which is the space of bounded infinite sequences of real numbers. The norm of any sequence is the greatest absolute value of any element of the sequence. –  Ross Millikan Sep 21 '12 at 0:19

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I will denote your space of all bounded real sequences by $X$ and let $M \subset X$ be as in your post. Also, for the sake of simplicity later, let's denote the $i$-th term of the sequence $z$ by $z(i)$ (instead of the usual $z_i$). Since you specifically asked about the norm, it is given by $$\|z\|_X = \sup_{i \in \mathbb{N}} |z(i)|$$

Let $x$ be the sequence defined by $x(n) = 1/n$ and let $x_m$ be the sequence defined by truncating the sequence $x$ after the first $m$ terms. That is, $x_m(n) = 1/n$ if $n \le m$ and $x_m(n) = 0$ otherwise.

Then we have that $x_m \in M$ for all $m$ and $$\|x - x_m\|_X = \sup_{n \in \mathbb{N}}|x(n) - x_m(n)| = 1/(m+1)$$

so that $\|x - x_m\|_X \to 0$ and hence $x_m \to x$ in $X$. So we have shown that $x_m$ is a convergent sequence in $X$. Therefore, it is also a Cauchy sequence. Moreover, the all the $x_m$ lie in $M$, but the limit $x$ is obviously not in $M$ (it is never $0$). Hence $x_m$ is a Cauchy sequence in $M$ which does not converge in $M$ (even though it converges in $X$).

In a sense, this is the only type of example you can come up with. Since $X$ is complete, a subset $M$ will be complete if it is closed which would mean, of course, that every Cauchy sequence in $M$ will converge in $M$. So you have to show that $M$ is not closed, which we did above by finding a convergent sequence of terms in $M$ whose limit is not in $M$.

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