0
$\begingroup$

This is the point of my confusion: Let $M$ be a manifold and $K$ an open subset of $M$, then $K$ is a submanifold of $M$. Submanifold are manifolds themselves, so we can speak about dimensions of $M$ and $K$ respectively. I need: do $M$ and $K$ have the same dimension? Some examples show that they do: let's say interval is not open in $R^2$, so all open subsets of $R^2$ are two-dimensional.

Why I am confused is how we can prove the existence of outward normal, which exists it is enough to say about $\Omega \subset R^n$. I then know that $\Omega$ has the same local vectors as $R^n$, and outward normal is locally normal to all vectors, to all $n$ independent vectors which is impossible in $R^n$. But, outward normal does exist... Where do I make a mistake??

1 Answers 1

0

We put outward normal on $\partial \Omega$, which has one dimension less, allowing that to happen. And them $\Omega$ MUST be the same dimension, or dimension $n$, because $\partial \Omega$ has a dimension exactly one less then $\Omega$, and if $\Omega$ had dimension $k < n$ then $\partial \Omega$ would have dimension $k-1$, allowing to have normal space of dimension $n-k+1$, which is more than 1, which means we have that outward normal isn't well defined.

  • 0
    Sometimes trying to explain your question to a friend who knows no mathematics at all solves problem for you, I saw a solution when he told me "tangent is on boundary of circle, not on circle". He could be a hell of a geometrist.... – 2017-02-14
  • 0
    Yes, this is the idea. $\Omega$ cannot be an arbitrary open set, though - the topological boundary of an open submanifold is not necessarily a submanifold. We typically would require $\Omega$ to be a "domain with $C^1$ boundary". – 2017-02-14