# Why the whole theory about differentiable manifolds is based on open sets?

I only have studied basic topology, which means i haven't studied any about differentiable manifolds. I just skimmed pages on wikipedia.

Here is a simple illustration on a basic situation.

Let $E$ be a subset of $\mathbb{K}$ and $(V,\| \cdot \|)$ be a normed space over $\mathbb{K}$ and $f:E\rightarrow V$ be a function.

It is only neccessary to check whether a point $x$ of $E$ is a limit point of $E$ to make a sense to say that "$f$ is differentiable at $x$.

So that we can define the differentiation of a function whose domain is such as a closed disk in $\mathbb{C}$ or the Cantor set.

For this reason, i guess the concept of differentiable manifold does not capture all the properties from which the concept of differentiation comes (i.e. $\mathbb{K}$), since the theory of differentiable manifolds only consider open sets.

I guess that if we consider all these cases together, then may the theory look messy. So i think all the concepts of differentiable manifolds are indeed substitutes to this broad consideration.

Am i thinking correctly or am i actually not capturing the intrinsic concept of differentiation?

While I feel the generalization of integration to a locally compact Hausdorff space is very natural, i kind of feel that the theory of abstract differentiation is a bit artificial.

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Note that there is the related concept of manifold with boundary, that are locally modeled on the open sets of the half space $\{ (x_1,\dots,x_n)\ |\ x_n\geqslant 0\ \}$. These open sets, as soon as they contains points of the form $(x_1,\dots,x_{n-1},0)$ are not open in $\mathbb R^n$. So we have some freedom on the notion of open sets when considering manifolds. – Taladris Aug 24 '14 at 12:41
There is also a notion of manifolds with corners, modeled on the open sets of $({\mathbb R}_{\geqslant 0})^n$. – Taladris Aug 24 '14 at 12:43