By Manifolds I will just mean smooth manifolds .
If $C$ is any closed subset of $\Bbb R^k$, then show that $\exists$ a submanifold $X$ of $\Bbb R^{k+1}$ such that $X \cap \Bbb R^k = C$ .
I here assume the result from Analysis that $\exists$ a smooth map $g : \Bbb R^k \to \Bbb R$ such that $C = g^{-1}(0)$
Now if $g$ is regular at all points of $C$ (i.e. $dg_x : \Bbb R^k \to \Bbb R$ is surjective $\forall x \in C$ ) then, $C$ would have been a submanifold of $\Bbb R^k$ itself and hence $X$ could have just been the canonical inclusion of $C$ in $\Bbb R^{k+1}$ !
This is clearly not the case always!
So I'm trying to use $g$ to get a transversality argument.
Like I had thought of $h : \Bbb R^{k+1} \to \Bbb R$ defined by, $h(x_1,\dots,x_k,x_{k+1})=g(x_1,\dots,x_k) + x_{k+1}$ then clearly $h$ is transversal ro $\Bbb R$ but I am unable to keep track of $C$ as nothing can be assumed about the behaviour of $g$ outside $C$ (except the fact that it is of course smooth there) .
Okay, so given our closed set $C \subset \mathbb{R}^k$, and taking a smooth function $g\colon \mathbb{R}^k\to\mathbb{R}$ with $C = g^{-1}(0)$, what if we take $X$ to be the graph of $g$, namely
$$X := \{ \left(x_1,\dotsc,x_k,g(x_1,\dotsc,x_k)\right) \mid (x_1,\dotsc,x_k) \in \mathbb{R}^k\}.$$
Since $g$ is smooth, it should be easy to check that $X$ is a $k$-dimensional embedded submanifold of $\mathbb{R}^{k+1}$. Then the fact that $X\cap \mathbb{R}^k = C$ comes for free from our assumption on $g$.