Name for a subset of real space being nowhere a manifold with boundary

[disregardthat]

I was wondering if anyone knew of a name for such a set, namely a subset $S \subseteq \mathbb{R}^n$ which at every point $x \in S$ there exists no open subset $U$ of $\mathbb{R}^n$ containing $x$ such that $S \cap U$ is homeomorphic to either $\mathbb{R}^m$ or the half-space $$\mathbb{H}^m = \{(y_1,...,y_m) \in \mathbb{R}^m : y_m \geq 0\}$$ for any integer $m \geq 0$. Of course, any set for which such open sets $U$ exists for every $x$ is called an embedded manifold with boundary. I'm looking for the opposite notion, in a sense.

 

[Office_Shredder]

The complement of a dense set?

 

[disregardthat]

I've corrected my question a bit, it now more accurately reflects the title. The complement of a dense set for the original question (where $m$ had the fixed value $n$) could very well be the correct answer. I can't prove it right now, but I'll look into it. For general $m$ however, it's not the right notion. For example, I'd like the graph of a nowhere continuous function to fit the bill, but not the graph of a continuous function.

EDIT: Upon some further reflection it dawns on me that the half-space condition is unecessary. If a space is isomorphic to the half-space locally around any point, then it is necessarily isomorphic to euclidean space at a nearby point. Hence the half-space condition can be dropped entirely without changing the question.

 

[Office_Shredder]

I missed the $m \neq n$ idea, sorry.

 

[disregardthat]

I missed the $m \neq n$ idea, sorry.

It was my fault, I didn't include it in the original formulation

 

[mfb]

It still needs to be the complement of a dense set I think, but not every complement works (because {0} is the complement of a dense set for example).