$(V, ||.||)$ normed space and define $d(x,y) = ||x|| + ||y|| $ if $ x \neq y$ and as $0$ if $x=y$. Convergent sequences in $(V,d)$? Complete? Let $(V, ||.||)$ be a normed space and define $d(x,y) = ||x|| + ||y|| $ if $ x \neq y$ and as $0$ if $x=y$. Describe all convergent sequences in $(V,d)$.
I'm sure that all eventually sequences are convergent sequences as in this case $d(x_n,x) \to 0$.
I'm not really sure how to prove these are the only convergent sequences in this space / whether these are in fact all the convergent sequences in this space.
Also, what is the general approach for finding all convergent sequences in a metric space?
How do I go about proving this is complete?
Thanks!
 A: If $x_n \to x$ in $d$, there are two cases:
$x=0$. Then $d(x_n,x)= \|x_n\| + \|x\| = \|x_n\|\to 0$ and so $x_n \to x$ under $\|.\|$. So $(V,\|.\|)$ and $(V,d)$ have the same convergent sequences with limit $0$.
$x \neq 0$. Then also $d(x_n,x) \to 0$ but $d(x_n,x) \ge \|x\| >0$ so this can never happen unless $x_n = x$ for a tail of the sequence (because then the other clause of the definition of $d$ kicks in). So a sequence can only converge in $d$ to a $x \neq 0$ if it's eventually constant with value $x$.
This completely characterises the convergence in $(V,d)$: if a given sequence is not eventually constant (then it has the obvious limit), it converges (to $0$) iff it already did under the norm.
Similar considerations show that $(V,d)$ is complete: if $(x_n)$ is Cauchy $d(x_n, x_m) = \|x_n\| + \x_m\|$ must tend to $0$ for both $n,m \to \infty$. So if all $x_n$ are bounded away from $0$, an injective sequence cannot be Cauchy, so $x_n \to 0$ follows by necessity.
A: $d(x_n, x)\to 0$ iff $\|x_n\| +\|x\|\to 0$ iff $\|x_n\|\to 0 \ \& \ \|x\|\to 0$ iff $x_n\overset{\tiny\| \cdot \|}{\to} 0 \ \& \ x=0$
A: Let $x_n \to x$, by definition: $\forall \varepsilon > 0 \ \exists \ N \in \mathbb{N}: \ n \geqslant N \Rightarrow ||x||+||x_n||=d(x_n,x)<\varepsilon.$
Note that $|| x || \leqslant ||x|| + ||x_n||< \varepsilon. \ $ Since $\varepsilon$ is arbitrary, we must have $||x||=0 \Rightarrow x=0.$ If we substitute $||x||=0$ for the property $x_n \to x$, we have:
$\forall \varepsilon>0 \ \exists \ N \in \mathbb{N}: \ n \geqslant N \Rightarrow ||x_n||<\varepsilon $. Since $||x_n||=||x_n-0||$, it's easy to see that convergence wrt $d$ implies convergence to $0$ with the metric $d'(x,y)=||x-y|| $, the standard metric derived from norm.
On the other hand any sequence converging to $0$ wrt the metric coming from the norm, clearly converges with respect to $d$, it's just a matter of checking the definitions.
Completeness:
if $x_n$ is Cauchy, $\forall \varepsilon > 0, \ \exists \ N \in \mathbb{N}: n,m > N \Rightarrow d(x_n,x_m)<\varepsilon$.
Note that $||x_n|| \leqslant d(x_n,x_m) = ||x_n||+||x_m|| < \varepsilon$. Thus $x_n \to 0$ with respect to the metric coming from $|| \cdot ||$.
Any sequence $x_n \to 0$ wrt to $||\cdot ||$, is Cauchy wtr to $d$.
