Is this proof valid for showing that this normed vector space is a Banach space? A student suggested a way of proving that the following space defines a Banach space, but I'm not sure it can be done in this way.
Let $\mathbb{F}=\mathbb{C}$ or $\mathbb{R}$. Consider so the usual vector space $\mathbb{F}^2$ with the norm defined by $\| (x,y) \| = |x| + |y| $. To show that this defines a Banach space, define a Cauchy sequence $(x_n,y_n)$ in $\mathbb{F}^2$, note that each term in the vector converges in $\mathbb{F}$ to say $x_0$ and $y_0$ by the general Cauchy principle. These limits are both in $\mathbb{F}$, so $(x_n,y_n) \rightarrow (x_0,y_0) \in \mathbb{F}^2$
So my intuition after reading this, is that this is not a valid proof, as the (induced) metric defines convergence, and here the metric has been substituted for another when checking for convergence in $\mathbb{F}$. Am I right in saying this? If so, can anyone come up with counter-examples for when this argument holds, but the space is not complete?
 A: Every finite-dimensional normed space of dimension $n$ over $\mathbb{F}$ is isomorphic to $\mathbb{F}^n$, and hence complete (assuming $\mathbb{F}=\mathbb{C}$ or $\mathbb{R}$).
Your student's argument works! The space that this norm defines (in $n$ dimensions) is somtimes called $l_1^n$ (which is complete). You can also apply the same norm (infinite sum) to the space of sequences $(a_n)$ such that $\sum |a_n| < \infty$, called $l_1$ (which is also complete, although you'll need a more sophisticated argument).
Here is the full argument written out for clarity:
Suppose $(z_n) \in l^2_1$ is Cauchy. Write $z_n=(x_n,y_n)$. 
Note that $|x_n| \leq \|z_n\|= |x_n| + |y_n|$, so $(x_n)$ is a Cauchy sequence in $\mathbb{F}$. Similarly, $(y_n)$ is Cauchy in $\mathbb{F}$. Hence $x_n \to x$, $y_n \to y$ for some $x,y \in \mathbb{F}$, using completeness of $\mathbb{F}$. It's easy to see that this implies that $\|z_n - (x,y)\| \to 0$, and so $z_n \to (x,y) \in \mathbb{F}^2$.
Note that you do switch norms in the proof, and then switch back. However this is valid, as you are invoking the definition of the norm you put on the product space.
A: In finite dimensions, all norms are equivalent. 
(For example, see Any two norms on finite dimensional space are equivalent.)
This means the following: if $\Vert \cdot \Vert_1$ and $\Vert \cdot \Vert_2$ are the two norms on the space $X$, there are constants $a>0$ and $b>0$ such that
$$
a \Vert x \Vert_1 \leq \Vert x \Vert_2 \leq b \Vert x \Vert_1,
$$
for every $x \in X$. If you have this, then it is easy to show that $\Vert x_n -y \Vert_1 \to 0$ if and only if $\Vert x_n -y \Vert_2 \to 0$. So changing the norms will not change whether the space is complete (in the finite dimensional case).
 So if you add this fact to your proof, then this proof works in the finite dimensional case. In particular, it works for $\mathbb F^2$. Without using this fact (or a similar one), this proof is incomplete.
If the space is infinite-dimensional, then you are right to say that changing the norm can cause problems. For example, consider $C[0,1]$ (the continuous functions on the interval $[0,1]$) with the following two norms:
$$
\Vert f \Vert_\infty = \sup_{x \in [0,1]} |f(x)|;
$$
$$
\Vert f \Vert_1=\int_0^1 |f(x)| \ \mathrm dx.
$$
The space $(C[0,1],\Vert \cdot \Vert_\infty)$ is a Banach space. (For example, see the answers to How to show that $C=C[0,1]$ is a Banach space.) However the space $(C[0,1],\Vert \cdot \Vert_1)$ is not a Banach space (for example, Showing $(C[0,1], d_1)$ is not a complete metric space). So here the proof would not work. Changing the norms does affect whether or not the space is complete.
