This question was in my manifolds notes which I borrowed from a senior and I was unable to prove the result.
Let S be an k-dimensional submanifold of an n-manifold M with k<n. Show that , for every point $p\in S$, there exists an open set $U\subset M$ containing p, and a smooth function $f: U \to \mathbb{R}^{n-k}$ such that $f^{-1}(0) =U \cap S$.
For every $p\in S$ there will always exists an open set $U \in M$ containing p but I have to find a specific U that also satisfies the condition on the smooth function.
A subset S of M is called a k-dimensional submanifold of M if the following holds: If $p\in S$ there exists a chart $(U,\phi)$ containing P st $\phi(U\cap S)$ is an open subset of $\mathbb{R}^k$.
But How to use it to find such a function. I am really sorry but I am unable any way to prove it. I couldn't add much to my attempt and thoughts because I am struck. Kindly don't close this question.
Please shed some light on this!
Since $S$ is a $k$ dimensional submanifold of the $n$ dimensional manifold $M$, for any $p\in S$, there exists an open subset $U \ni p \subset M$ and a chart $\phi\colon U \to V\subset \Bbb R^n$ (with $V$ open) such that $\phi(U\cap S) = V \cap (\Bbb R^k \times \{0\}^{n-k})$. Consider the linear projection $\pi\colon \Bbb R^n \to \Bbb R^{n-k}$ defined by $\pi(x_1,\ldots,x_n) = (x_{k+1},\ldots,x_n)$, and define $f \colon U \to \Bbb R^{n-k}$ by $f = \pi\circ \phi$. Then $f$ is a solution.
The existence of such a chart is exactly the definition you gave. The embedding $\Bbb R^k \subset \Bbb R^n$ isn't explicit in your definition, but it surely is of the form $\Bbb R^k\times \{0\}^{n-k} \subset \Bbb R^n$.