this seemingly innocent question has been bugging me for quite a while. Lets give a minimal example:
The unit circle S has no boundary considered as a manifold (all points have neighborhoods homeomorphic to R). This coincides with the notion of the topological boundary when using the topology induced by the charts (as the entire circle is then open, its interior is its closure). Moreover, this topology coincides with the topology induced onto the circle as a subspace of R²
Now we turn to the unit interval [0,1]. Considered as a manifold, its boundary are 0 and 1 as these points are only homeomorphic to the semiopen intervals. But of course, any topology that considers [0,1] as its own space and not as a subset of R must have [0,1] open, which leads to the vanishing of its boundary in the topological sense.
So i have failed to construct any topology on [0,1] in a way thats consistent with the one on the circle that would allow me to say 'the circle has no boundary, the unit interval does', unless i accept that we always talk about the manifold boundary, which must be different from the boundary of the manifold in a topological sense (in which case, using the same word for those two concepts does not exactly feel optimal to me, cosidering differential geometry and topology are so closely related).
can any one confirm or explain me where i went wrong here? thanks in advance