- #1
kent davidge
- 933
- 56
Whitney's theorem tell us that a m-dimensional manifold embedds in ##\mathbb{R}^{2m}##, but does not tells us if it embedds in ##\mathbb{R}^n, n < 2m##. In special, I wanted to know whether the circle embedds in ##\mathbb{R}##. For this, I tried constructing a function ##f: \mathbb{S}^1 \longrightarrow \mathbb{R}## given by ##f(\cos\theta, \sin\theta) = \theta## for ##(\cos\theta, \sin\theta) \in \mathbb{S}^1##. The image is ##f(\mathbb{S}^1) = (0, 2 \pi]## if ##\mathbb{S}^1## is defined as usual as a subset of ##\mathbb{R}^2## with ##\theta## on that interval.
Such function ##f## seems to be the more natural mapping I can think of, for its inverse is how ##\mathbb{S}^1## is typically defined. The problem is that ##(0, 2 \pi]## is not open in ##\mathbb{R}## (usual topology) whereas ##\mathbb{S}^1## is open in itself (subspace topology). So should I conclude that ##\mathbb{S}^1## does not embedd in ##\mathbb{R}##?
Such function ##f## seems to be the more natural mapping I can think of, for its inverse is how ##\mathbb{S}^1## is typically defined. The problem is that ##(0, 2 \pi]## is not open in ##\mathbb{R}## (usual topology) whereas ##\mathbb{S}^1## is open in itself (subspace topology). So should I conclude that ##\mathbb{S}^1## does not embedd in ##\mathbb{R}##?