On what structural level should one think of $R^3$ Say i have the space $R^3$. Is it ok to think of this space only as a set therefore to think that it has no structure (only cardinality), or i should not think of the space only as a set?
I understand $R^3$ is a set, my question is if i can view it only as set and therefore with no other structure.
Thanks 
 A: This is the type of question that category theory was invented to answer.  In category theory, you have objects and morphisms; for many frequently used categories, the objects are sets with some additional structure, and the morphisms are functions that preserve that additional structure.  So how you think of $\mathbb R^3$ depends on what you are trying to do with it, or as I think of it, what kind of math you are doing.
So:


*

*in the category of sets, $\mathbb R^3$ is as you say, an uncountably infinite set with no particular structure beyond its cardinality.

*in the category of topological spaces, $\mathbb R^3$ comes with an implied topology, a normal topology, unless specified otherwise.  In this topology $\mathbb R^3$ is path-connected, contractible, normal, can be described as the countable union of compact sets, and many other properties.  Morphisms are continuous functions from $\mathbb R^3$ to other topological sets.  Even without discussing smooth maps or metrics, in topology, $\mathbb R^3$ has so much specialized structure you can spend ages talking about it.

*As a metric space, $\mathbb R^3$ is usually assumed to have the Euclidean metric, again unless specified otherwise.  With a topology coming from its metric, $\mathbb R^3$ can be distinguished from $\mathbb R$ easily, in that it does not admit an ordering compatible with that metric (in this sense: why is there no order in metric spaces? ).  Second, it can be distinguished from all other $\mathbb R^n, n \ne 3$
because the set of points distance 1 from the origin is a 2-sphere.
The obvious morphisms in a category of metric spaces would then be isometries, or isometric embeddings, and the former can be related to isometries of this 2-sphere; but stepping away from strictly considering morphisms, there is a lot more one considers special about $\mathbb R^3$ as a metric space.

*in the category of vector spaces, $\mathbb R^3$ is understood to be a 3-dimensional real vector space, and morphisms are linear maps from it to other real vector spaces.  This is to me the best reason to even call it $\mathbb R^3$ in the first place: i.e. when you say $\mathbb R^3$ you are signaling to me first and foremost that you mean a vector space.  In addition to purely linear maps one often uses the Euclidean metric on $\mathbb R^3$ just as with any other $\mathbb R^n$, and the cross-product on $\mathbb R^3$ in particular.

*as a smooth manifold, $\mathbb R^3$ is a topological space, a metric space, and has a tangent vector space (also $\mathbb R^3$, but this time considered as a vector space first) at each point, and because its tangent spaces are $\mathbb R^3$ it is called a 3-manifold.  One may construct other smooth manifolds out of $\mathbb R^3$ using smooth bijective functions from open sets in $\mathbb R^3$ to others.  (Algebraic geometers will tell you about how sheaves and $\mathbb C^3$ are better than this.)
There are many more things one could say.  My advice is, just don't get stuck only in pure set theory and lose sight of what can be done with it as a "space", or go too far the other direction (as many do) and assume every structure one can impose on $\mathbb R^3$ need always be considered.
A: 
Is it ok to think of this space only as a set therefore to think that it has no structure (only cardinality)

Yes, it's okay to do this.
Say I give you a set $A$ with elements $a,b,c \in A$ and no other elements (so $A = \{a, b, c\}$ ). The only way you can think of this set is as a set with three elements, because you don't know what the elements are (the names don't matter, it could have been $Z = \{ w, y, cherry \}$ ).
But now suppose I told you that $a$ is 1, $b$ is 2, and $c$ is 3. Before, you didn't know that $A$ had an order on it, now you do. But $A$ is the exact same set as it was before.
To expand further on the order aspect, you can set any order you want on the elements of $A$: $a < b < c$, $b < a < c$, etc. It doesn't even have to be total, you can say $a \le b$ and $a \le c$ but $b \nleq c$. And yes, this is even though $a = 1$, $b = 2$, $c = 3$. Just because it's different from the "normal" ordering we think of for numbers doesn't make it wrong.
