Manifold with the same dimension as $\mathbb{R}^n$

Question

I am studying manifolds at a basic level, and I was wondering if, when a set has the same dimension as the space where it is defined (take, for instance, $\{ (x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 < 1 \}$), it neccesarily needs to be open to be a manifold. For example, I know that the previous example is it, but if one replaces $<$ by $\leq$, then it is closed and I know it is not a manifold. In this case, since the dimension of both $\mathbb{R}^3$ (the space in which is embedded) and the disc of radius $1$ are the same ($3$), is the fact that one is open and the other not what makes the difference? I ask this because for a 2-dimensional set in $\mathbb{R}^3$, it can be closed and still be a manifold. Take for example $x^2 + y^2 + z^2 = 1$.

Answer 1

It is true that if a subset of $\mathbb R^n$ is an $n$-dimensional manifold then that subset must be open. This is a deep theorem of algebraic topology, known as the Invariance of Domain Theorem.