This is probably a very silly question but it bothers me for some time.
We define a metric space $X$ to be complete if every Cauchy sequence in $X$ converges to some point in $X$.
But any metric space $X$ is closed in itself, so every convergent sequence in $X$ has its limit in $X$. From that I would say that when we check if $X$ is complete, it suffices to check that any Cauchy sequence in $X$ converges because if it converges, then it will converge in $X$.
However...
consider for example the space of continuous functions on $[0,1]$, $C([0,1],\mathbb R)$ with $\|\cdot\|_1$. Take $(f_n)_{n\geq1}$ given by
$$f_n(x)=\begin{cases} 0,&0\leq x\leq\frac12\\ n\left(x-\frac12\right),&\frac 12<x\leq \frac 12+\frac1n\\ 1,&\frac12+\frac1n<x\leq 1\;. \end{cases}$$
This is continuous for each $n\geq 1$ and so in $C([a,b],\mathbb R)$. We can show that it is a Cauchy sequence and that it converges in the $\|\cdot\|_1$ norm to $f$ defined by $f(x)=1$ when $x\in(1/2,1]$ and $f(x)=0$ otherwise.
If my reasoning above were correct, then we would stop now and say that $C([0,1],\mathbb R)$ is complete, which of course it isn't because the $f$ we found above is not continuous.
I think that the problem has to do with the universe we are in. Because if we consider $C[0,1]$ as just a subset of $R[0,1]$ (where this is the set of Riemann integrable functions on $[0,1]$), then of course it is not closed in $R[0,1]$. But we are not doing that, we are just referring to $C[0,1]$, or more generally to the metric space $X$ and assume that $X$ is the "universe."
So what am I missing here?
Brian's comments are spot on. Let me add a little (too long for a comment): Properties like "Cauchy" or "convergent" depend on the metric space you're living in. It doesn't make sense to say that the sequence $(3, 3.1, 3.14, . . . )$ converges or doesn't converge - you have to say what metric space you're in. In $\mathbb{R}$ with the usual metric, this does converge; in $\mathbb{Q}$ with the usual metric, it doesn't. And the sequence isn't even Cauchy in $\mathbb{R}$ with the discrete metric $d(x, y)=1$!
Usually, of course, we just say things like "$S$ is Cauchy" - but this is a quirk of natural language. Technically, we should say what space we're living in, but we don't when it's clear from context.
There is, however, a useful sense in which Cauchy-ness is "more absolute" than convergence: namely, if $S$ is a sequence in a metric space $M$, and $N$ is a larger metric space, then $S$ is Cauchy in $M$ iff $S$ is Cauchy in $N$. This is not true for convergence. In particular, Cauchy-ness depends only on the distances between the points in the sequence itself; convergence depends on the "rest" of the space.
There is a precise sense in which every Cauchy sequence in a metric space $M$ is convergent "somewhere" - namely, in the completion of $M$! But this shift in the underlying space is important.