When talking about a normed vector space, does it's metric always need to be the induced one? The title basically says it all. If we have a normed vector space, is it possible to work with the space as a metric space with a different metric than the induced one?
So if the space is $(X,||\ ||)$, what happens if we starting working with a metric $d(x,y) \neq ||x-y||$? 
I'm asking because I came across the following theorem:
if $||\ ||$ is a norm, then $||\ ||:X\rightarrow\mathbb R$ is continuous
But atleast one definition of continuity involves the metric defined on a space. So I would assume that for the theorem to be true, there has to be a certain 'compatibility' between the chosen metric, and the chosen norm. However, if this is indeed true, it is never explicitly mentioned. The text I'm reading just mentions that $d(x,y) = ||x-y||$ is a possible metric, not that we always have to chose it in order for any of the results to be true.
 A: A normed space comes with a natural metric, $d(x,y) = ||x-y||$. This natural metric gives us a topology $\mathcal{T}$, and a topology gives us a notion what a continuous function on that space is. Unless otherwise stated, this is the topology a normed space.
Of course it's possible to consider other metrics $d'$ on $X$, e.g. other ones that induce the same topology $\mathcal{T}$, like $d'(x,y) = 3||x-y||$, or $d'(x,y) = \min(||x-y||,1)$. This is not normally done, though, as the metric from the norm has special nice properties that these other metrics might not have. Using a metric that induces a totally different topology is not usually done, considering other non-metric topologies is (weak topology, e.g.). 
A: The metric is purely your freedom to choose, you can even work on a normed space endowed with a discrete metric if that would be useful for some reason. Such examples often pop up in measure theory and functional analysis: measures naturally form a vector space, and the total variation norm is quite useful in many problems related to measures. Yet, there are many many metrics over measures each of them useful in relevant application field, and the metric induced by the total variation norm is just one of them.
Yet you right in the following: unless contrary is explicitly mentioned, when working on normed spaces and seeing notions from topology (e.g. continuity) or metric (e.g. completeness) it's very much likely the author means the metric and topology induced by the norm.
A: This is a matter of linguistic efficiency. 
You can equip a normed space with a metric that is not the one induced by the norm.
And then you can equip it with a topology that is not the one induced by the metric. You could even define a new addition that is not the vector addition in $X$. Or you could decide to pick a specific nonzero element of $X$ and call it $0$. But please never do anyting of this without mentioning it very specifically! 
So a corresponding statement should be written as

"If $\|\,\|$ is a norm on $X$ and $d$ is a metric on $X$, then ... is continuous with respect to the topology induced by the metric." 

As it is not written like that, the given statement is understood to be interpreted as 

"If $\|\,\|$ is a normon $X$, then the map $\|\,\|\colon X\to \mathbb R$ is continuous with respect to the topology induced by $\|\,\|$ on $X$ and the standard topology on $\mathbb R$".

If we have a vector space $X$ and use an element of $X$ called $0$ without further comment, then this $0$ is the additive neutral of $X$. If we write a symbol $+$ between to elements of $X$ without comment then this is the vector addition in $X$. If we talk about the topology of a metric space (e.g. by speaking about the continuity of a map) and do not mention otherwise, then the topology is  the topology induced by the metric. And finally, if we talk a bout a metric in a normed space and do not mention otherwise then the metric is the one induced by the norm.
