The boundary must have no boundary...

Question

It is more complicated than I thought. Let's make the hypothetical statement simpler.

Suppose that an $n-1$ dimensional hypersurface is the boundary of a connected $\mathbb R^n$open set, then it must be a hypersurface without boundary.

Basically, I would like to know, "can a manifold with boundary in the plane be the topological boundary of a $R^2$ open set?"

Are there some simple proofs for this simple statement?

Old question: The boundary of a "connected" $R^n$ open set must be a $(n-1)$ dimensional hypersurface without boundary. This statement seems "intuitively correct" in my humble sense. Which theorem shall I cite?

Answer 1

Edit to answer the question in its final form:

Basically, I would like to know, "can a manifold with boundary in the plane be the topological boundary of a $R^2$ open set?"

Yes. Consider the unit interval on the $x$-axis.

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It's more complicated than that, whatever you mean by "hypersurface without boundary":

A set is the boundary of some open set if and only if it is closed and nowhere dense.

https://en.wikipedia.org/wiki/Boundary_(topology)

There are many such sets that are not hypersurfaces. Consider the boundary of the open set of all points in the plane both of whose coordinates are not integers. The boundary is the set of grid lines: points with at least one integer coordinate.

There are far more complex examples. The Mandelbrot set is closed, so its complement is open. The boundary of that open set is quite unlike any hypersurface ...