Check for compactness and connectedness of subspace P = $\{(x, y, z)\in \mathbb{R}^3 : z = x^2+y^2+1\}$ I have got this problem in my exam: Check for compactness and connectedness of subspace P = $\{(x, y, z)\in \mathbb{R}^3 : z = x^2+y^2+1\}$
I think given subspace P is closed and bounded subset of $\mathbb{R}^3$ so it shoould be compact. But I am not sure with this.
How to check for connectedness?
I am stucked on this problem. I need help to understand how to solve such kind of problems? 
Edit: I am not that much good in topology. I have started studying this subject. I need answer with little more explanation. Please take this pain.
Thanks for helping me
 A: The first step is to visualize the surface. This is what provides motivation for the solution below. It looks like a cone.
The surface can't be compact, because it is unbounded. We can take $x$ and $y$ to be as large as we want and get a corresponding point on the surface. 
The surface is connected. To see this, recall that path connectedness implies connectedness. Fix a point $(x_0,y_0,x_0^2+y_0^2+1)$ on the surface, and look at the path $f(t):[0,1]\rightarrow P$ given by $(tx_0,ty_0,t^2(x_0^2+y_0^2)+1)$. This connects every point to $(0,0,1)$. 
A: Consider the set $\{(t,0,t^2+1)\}_{t \in \mathbb{R}}$. Is this set contained in $P$ and is it bounded?
It is straightforward to show that $P$ is path connected. Suppose $(x_i,y_i,z_i) \in P$, $i = 0,1$. Define the path: $x(t) = x_0+t(x_1-x_0)$, $y(t) = y_0+t(y_1-y_0)$, $z(t) = 1+x^2(t)+y^2(t)$. Then $(x(t),y(t),z(t)) \in P$ with $t \in [0,1]$, and it connects the two points.
A: Visualization: Consider the equation $z=x^2+y^2+1 \Leftrightarrow x^2+y^2=z-1$ 
Note that LHS is always non-negative $\forall\  x,y \in \mathbb R$ so RHS is always non-negative but RHS is non-negative iff $z\geq 1$. At $z=1$ it represents the point $(0,0,1)\in \mathbb R^3$ and at $z=c>1$ it represents the circle $x^2+y^2=c-1$. By taking arbitrary $c>1$ you are generating a circle with center  $(0,0,c)$ and radius $\sqrt {c-1}$. Hence you are generating a cone.
With this visualization you can get that it can't be compact but connected.
ADD: To show path connected it need only to show that any two point is connected.Take $(x,y,z),(a,b,c)\in P$. Connect like this $(x,y,z)\rightarrow (0,0,1)\rightarrow (a,b,c)$. No need to do hardcore computation.
