Let $M$ be an embedded $k$-submanifold of $\mathbf R^n$ and $p$ be a point in $M$. For $\epsilon>0$, let $S_\epsilon(p)$ denote the sphere in $\mathbf R^n$ of radius $\epsilon$ having its center at $p$. So $S_\epsilon(p)$ is itslef an embedded $(n-1)$-submanifold of $\mathbf R^n$.
Is the following true?
For $\epsilon$ small enough, $S_\epsilon(p)\cap M$ is diffeomorphic to a sphere of dimension $k-1$.
Intuitively, near $p$ the manifold $M$ looks like a flat $k$-plane, and intersecting $S_\epsilon(p)$ with a $k$-plane passing through $p$ yields a $(k-1)$-sphere.
The reason why I ask this is the following: On pg 17 of Milnor's Singular Points of Complex Hypersurfaces, the author states the following (Corollary 2.9).
(Not exact quote) Let $V$ be a real algebraic set and $x^0$ be a non-singular point in $V$. Every sufficiently small sphere $S_\epsilon$ centered at $x^0$ intersects $V$ in a smooth manifold.
Since the set of singular points form a closed set, we note that a small enough sphere centered at $x^0$ will not contain any singular point of $V$. It is already stated in the book that the set of non-singular points of an algebraic set form a manifold, each of whose components have the same dimension (See Theorem 2.3). Milnor does not use the qualification 'embedded' for a manifold. But after reading the reference Milnor has given for the proof of Theorem 2.3, I am convinced that the set of non-singular points form an embedded submanifold of the ambient space. Further, the proof Milnor has given makes use of the algebraic nature of $V$.
So in the above Milnor is stating something even weaker than what I am intuitively finding true in a more general setting. Or may be I am missing some subtle point.
EDIT: I am a bit embarrassed to write this. But I should. The thing is that the theorem Milnor states is not just for non-singular points but also for isolated singular points. The proof provided by Milnor uses the algebraic nature of the problem just to cover the case of the isolated singular point. At any rate, @Shinchishiro Nakamura has resolved the question I had asked.