Finding a cauchy sequence that does not converge on M 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? =)
 A: 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 $M$.
