Some books on mean curvature flow (e.g. Mantegazza, Ecker) state that an embedded hypersurface in $\mathbb{R}^{k+1}$ is orientable (Mantegazza page 3, Ecker page 110). In other words, they assume the existence of a globally defined (unit) normal field along the embedded hypersurface.
Does anyone know how to prove this? Isn't the Mobius strip able to be embedded in Euclidean space, yet it isn't orientable?
For some definitions, let $M$ be a smooth $k$-dimensional manifold. A smooth map $\varphi:M\rightarrow\mathbb{R}^{k+1}$ is a hypersurface (an immersion) if its differential is injective. It is an embedding if it is also a homeomorphism onto its image $\varphi(M)$. An immersed hypersurface is a subset $S\subset\mathbb{R}^{k+1}$ which is a $k$-dimensional manifold such that the inclusion map $\iota:S\hookrightarrow\mathbb{R}^{k+1}$ is an immersion. An embedded hypersurface is a subset $S\subset\mathbb{R}^{k+1}$ which is a $k$-dimensional manifold whose topology conicides with the subspace topology inherited from $\mathbb{R}^{k+1}$ and the inclusion is an embedding.
We can also characterise an embedded hypersurface using single-slice adapted charts: Each point $p\in S$ is contained in a coordinate neighborhood of a single-slice chart $(U,\mathsf{u})$ on $\mathbb{R}^{k+1}$ adapted to $S$ such that $U\cap S=\{(u^1,…,u^{k+1})\;|\;u^{k+1}=0\}$. The pair $(U\cap S,\mathsf{proj}_{\mathbb{R}^k}\circ\mathsf{u}|_{U\cap S})$ is then a chart for $S$ and as we range over $p\in S$ we get an atlas for $S$.
Best Answer
Theorem
Every closed embedded hypersurface of $S\subset \mathbb R^{k+1}$ is orientable.
Proof
The hypersurface $S$ is defined by family $(U_i,f_i)$ where the $U_i$'s are an open covering of $\mathbb R^{k+1}$ and the $f_i:U_i\to \mathbb R$ are $C^\infty$ functions subject the condition that there exist $C^\infty$ functions $g_{ij}:U_i\cap U_j\to \mathbb R^\star$ satisfying $f_i=g_{ij}f_j$ on $U_i\cap U_j$.
We then have $S\cap U_i=f_i^{-1}(0)$
Those $g_{ij}$ define a line bundle $L$ on $\mathbb R^{k+1}$, which is necessarily trivial like all line bundles on $\mathbb R^{k+1}$, since $\mathbb R^{k+1}$ is contractible.
But then the restriction $L|S$ of $L$ to $S$ is trivial too and since that restriction is the normal bundle of the embedding $S\hookrightarrow \mathbb R^{k+1}$, that normal bundle is trivial, which implies that $S$ is orientable.
[Note carefully that connectedness or compactness of $S$ is not assumed]
Edit
In this more recent answer I prove independently that the smooth hypersurface $S$ has a global equation, which implies that it is orientable.