Compactness and connectedness in $\Bbb R^3$ 
Consider the set $$A=\left\{ \begin{pmatrix} x\\y\\z\end{pmatrix} \in \Bbb R^3: z=x^2+y^2+1\right\} \subset \Bbb R^3$$
Prove of disprove: $A$ is connected and compact

The set $A$ is unbounded, since $(\sqrt{n},\sqrt{n},2n+1)^t\in A$ with $$d(\textbf{0},(\sqrt{n},\sqrt{n},2n+1))=\sqrt{2n+(2n+1)^2}>n$$ where $\textbf{0}$ denotes the origin in $\Bbb R^3$ and $n \in \Bbb N$. So $A$ is not compact.
Is this correct ? How about connectedness? Can I have a hint?

The continuous map $f : \mathbb{R}^2 \to \mathbb{R}^3, f(x,y) = (x,y,x^2+y^2+1)$, establishes a homeomorphism between $\mathbb{R}^2$ and $A = f(\mathbb{R}^2)$. If $p : \mathbb{R}^3 \to \mathbb{R}^2, p(x,y,z) = (x,y)$ denotes the (continuous!) projection, then $p \mid_A : A \to \mathbb{R}^2$ is the inverse homeomorphism. This shows that $A$ is connected but not compact.
