Is this function continuous at $\vec{0}$ Divide $\mathbb{R}^2$ into the following disjoint sets:
$\mathbb{R}^2 = B_{+} \cup B_{-} \cup C_{+} \cup C_{-} \cup L$,
where
$$B_{+} = \{ (x,y) \in \mathbb{R}^2 \mid y > x^2 \}$$
$$C_{+} = \{ (x,y) \in \mathbb{R}^2 \mid 0 < y \leq x^2 \}$$
$$B_{-} = \{ (x,y) \in \mathbb{R}^2 \mid y < - x^2 \}$$
$$C_{-} = \{ (x,y) \in \mathbb{R}^2 \mid 0 > y  \geq - x^2 \}$$
$$L = \{ (s,0) \mid s \in \mathbb{R} \}$$
Let the function $f : \mathbb{R}^2 \to \mathbb{R}$ be defined as 
$f(\vec{x}) = 
\left\{  \begin{array}{ll}
0,  & \mbox{ if } \vec{x} \in B_{+} \cup B_{-}  \\
1,  & \mbox{ if } \vec{x} \in C_{+} \cup C_{-}  \\
s,  & \mbox{ if } \vec{x} = (s,0) \in L.
\end{array}  \right.
$
I want to determine whether or not this function is continuous at the origin. I think it is not, because:
Let $\varepsilon < 1$ and $\vec{x}\in C_+\cup C_-$, then $\|f(x)\|=1 > \varepsilon$, for any $\delta$ where $\|x\|<\delta$.
Please let me know if this is correct.
 A: Your proof is kindof correct. I think what's in your head is definitely the right proof. However, it hasn't come down on paper just right, and some things are not in the order they should be. Most importantly, you must decide on $\delta$ before even talking about $\vec x$, since the whole point is that no matter what $\delta$ is, there exists a vector $\vec x$ with certain properties. Less important (not at all, really, it's just a convenience), you can choose a specific $\varepsilon$, and a specific $\vec x$, instead of leaving them arbitrary. Here is how I would write what you've tried to convey in that last line:

Set $\varepsilon = \frac12$. For any $\delta > 0$, if $\delta > 0.2$, set $\vec x = (0.1, 0.005)\in C_+$, and if $\delta \leq 0.2$, set $\vec x = (\delta/2, \delta^2/8)\in C_+$. This means that $f(\vec x) = 1$, and at the same time $|\vec 0 - \vec x| <\delta$. But $f(\vec 0) = 0$, so $|f(\vec 0) - f(\vec x)| = 1 >\frac12 = \varepsilon$. Thus $f$ is discontinuous at $\vec 0$.

