How to show that a set isn't a submanifold?

Question

I'm trying on my own to learn about manifolds and would like to know whether this is a right way or not.

I used following defintion:
Let $n,N \in \Bbb N$. A non-empty subset M $\in \Bbb R^N $ is called a n-dimensional submanifold of $\Bbb R^N$, if for each $x \in M$ there exist:
(i) an open area $U$ of $x \in M$
(ii) an embedding of $\Phi: D \to \Bbb R^N$ with $D \subset \Bbb R^n$ open, such that $\Phi(D)=U$

The exercise is:
Let $ K= \{(x,y,z) \in \Bbb R : x^2+y^2=z^2\}$ be a cone.
Show that
(i) $K$ isn't a submanifold of $\Bbb R^3$
(ii) $K\setminus\{0\}$ is a 2-dimensional submanifold of $\Bbb R^3$

My attempt:
(ii) Let f:$\Bbb R^3 \to \Bbb R$ be with $f\begin{pmatrix}x \\ y\\ z\\\end{pmatrix}$=$x^2+y^2-z^2$

Than $\nabla f\begin{pmatrix}x \\ y\\ z\\\end{pmatrix}$=$\begin{pmatrix}2x \\ 2y\\ -2z\\\end{pmatrix}$
If $\nabla f=0 \Rightarrow x=y=z=0 , so f(x,y,z) \neq0$ that's why $f'$ has rank 1 in all points with $f(x,y,z)=0$
With the preimage theorem $ K$ is a 3-1=2-dimensional submanifold in $\Bbb R^3$

(i) Here I'm not sure, what have I to do exactly ? My attempt:

Let $K$ be a submanifold of $\Bbb R^3$ of dimension $n \in \{0,1,2,3\}$ in $\Bbb R^3$ , than where would be an open area $U$ around $(0,0,0)$ and a smooth mapping $g: U \to \Bbb R^{3-n}$ with rank $\nabla f_{(0,0,0)}=3-n$ such that $K \cap U =f^{-1}(0). $
Would it be enough to do case differentiations ?

Answer 1

For (i), yes, you can use the preimage theorem, but you can see it more directly: removal of $(0, 0, 0)$ divides $K$ in to two disjoint subsets both of which map bijectively onto the open subset $\Bbb{R}^2 \mathop{\backslash} (0, 0)$ of $\Bbb{R}^2$. The inverses of the two bijections give you the (homeomorphic) embeddings of $\Bbb{R}^2$ into $K$ that you need.

For (ii), note that removal of $(0,0,0)$ disconnects any open neighbourhood $D$ of $(0,0,0)$ in $K$. But no open subset of $\Bbb{R}^2$ can be disconnected by removing a single point. (If this doesn't make sense then you need to share some more information with us about the books you are using.)

Answer 2

Sometimes it is better to think of the problem without trying any computation. Try to think of it this way; you know how the cone lies in $\mathbb{R}^3$. Think of the point $(0,0,0)\in K$. If $K$ was a manifold, you would be able to find a neighbourhood of $0$ that is diffeomorphic (in particular, homeomorphic) to an open disk in $\mathbb{R}^2$. Can you see why that would yield a contradiction? Hint: use connectedness!