On linear span and its topological closure in infinite dimension This is kinda of a philosophical question I guess. But are the elmements of the topological closure inside the linear space $X$ all the time? Or do they become apperent when we introduce the topology? And hence introduce the topolgy to control these elements of the space which are there but out of control when we only have algebraic structure. i.e (I think) is the linear space smaller then "same" with the topology? Lets assume we have a reasonable topology.
 A: Here's how I would put it. We start out with a large linear topological space $\Omega$. Within $\Omega$ we isolate a linear subspace $X$ with nice properties (perhaps $X$ is the linear span of countably many elements, for example), and it may or may not be the case that each point of $\Omega$ is a topological limit of elements in $X$. 
On the other hand, there is a situation where we can start with a metric space and form its "completion", which refers to taking equivalence classes of Cauchy sequences, and the completion then becomes a metric space (hence a topological space) in its own right, with the original space as a subspace. This may be what you're referring to. Then we are really introducing something larger. 
Completion is one example of a situation when we start out with a topological space and then form a larger topological space from the original one (another is compactification). But these are not to be confused with the topological closure, which is non-vacuous only in reference to a larger space $\Omega$ which we already know contains $X$.
One thing you may be confused about is that (as far as I can tell) it does not make sense to say that taking the closure "adds a topology". The closure makes sense only if there is already a topology present--only if $X$ is already a topological subspace of a larger topological space. (The larger space could be $X$ itself, but this makes the discussion vacuous.)
You should think of a topology as specifying which elements are close to other elements. In the case of a linear space, there is already some notion of which elements are close to others (those which have close coefficients), and so the topology that we choose will back up this notion of closeness (unless for some bizarre reason we want to proclaim that elements with close coefficients are not the elements that are close to each other).
A good example to keep in mind is when $X$ is the finite linear combinations of  trig functions on a compact interval $C$ and $\Omega$ is $L^2(C)$. In that case, the closure of $X$ is in fact all of $\Omega$, although $X \subsetneq \Omega$. In particular, the closure of $X$ includes all continuous functions on $C$.
A: One subtle difference between metric spaces and topological spaces is that the "completion of a topological space" is not a well-defined notion. 
Of course the closure is well-defined at the level of topological spaces. But unlike the completion, the closure is not really a unary operation, it is a binary operation: we take the closure of a set $A$ in an ambient space $X$. This $X$ could be $A$, but then the closure of $A$ is just $A$. For instance the closure of $(0,1)$ in itself is $(0,1)$; thus the situation might not be so geometrically intuitive if we don't have some assumptions about the ambient space.
But in metric spaces, we can talk about completion without an ambient space. For instance, in $L^2([0,\pi])$, we can take the finite linear combinations of $e^{inx}$ for integers $n$. The completion of this is all of $L^2([0,\pi])$. (This is a slightly weaker statement than "everything in $L^2([0,\pi])$ has a $L^2$-convergent Fourier series", which is also true.)
Notably, the completion of a metric space cannot be determined by looking only at the induced topology. For example, compare $\mathbb{R}$ with the standard metric and $\mathbb{R}$ with the metric $d(x,y)=|\arctan(x)-\arctan(y)|$. These both generate the standard topology, but in the latter, $x_n=n$ and $x_n=-n$ are Cauchy. Thus the completion of the latter contains a point "at $+\infty$" and a point "at $-\infty$".
This difference can get obscured when we work with an ambient space which is a complete metric space like $\mathbb{R}^n$. In this situation the completion and the closure coincide.
