# 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

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Is this a take-home exam, or did you just sit for a timed exam and are asking how to do a question you were stuck on? – Potato Jun 2 '12 at 18:13
Try drawing a picture of the surface $P$. – Grumpy Parsnip Jun 2 '12 at 18:15
@Potato Neither homework nor take home exam. It has been asked in one of entrance exam. If you need link i may provide you. – srijan Jun 2 '12 at 18:16
It's clearly unbounded because we can take $x$ and $y$ as large as we wish, so it can't be compact. – Potato Jun 2 '12 at 18:17
@srijan The surface is a cone. For fix $z$ the equation $x^2+y^2=z-1$ describes a circle. – azarel Jun 2 '12 at 18:20

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.

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ya given set is contained in P. I am confused how to see it geometrically. $t^2+1$ represents parabola .surely it is unbounded, But in 3d how to visualize this set? – srijan Jun 2 '12 at 18:31
I have same question that I asked from potato. How did you got this path. Is there anything else that i need to study? I always find difficulty to deal with this kind of problems where I find myself unable to visualize given surface. – srijan Jun 2 '12 at 18:41
Well, with these things there is no recipe. You need to look at the surface and guess. I often try just linear interpolation first (only works in convex sets, of course) and if that doesn't work, I try interpolating some of the coordinates, and figure out the formula for the remaining ones so they lie on the surface. But visualizing the surface in some way is a huge help. – copper.hat Jun 2 '12 at 19:01
Now i understand. your answer says the two points $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ are connected by the function defined by $f(t) = (x_0+t(x_1-x_0), y_0+t(y_1-y_0), 1+x^2(t)+y^2(t))$. with t = 0 and t = 1 will give both points. Am i correct?thanks – srijan Jun 2 '12 at 19:57
Correct (and $f(t) \in P$, $\forall t \in [0,1]$, of course). – copper.hat Jun 2 '12 at 21:18

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 $(0,0,1)$.

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How can I know that such path would exist? I need little concept. I could never imagine this path. Please tell me. Thnaks for your answer. – srijan Jun 2 '12 at 18:34
Proving connectedness from the definition is hard, so it is best to look for path connectedness. Once you see every point is path connected to the origin from visualization, it is not hard to write out that path. – Potato Jun 2 '12 at 18:39
Why this point (0,0,1)? – srijan Jun 2 '12 at 18:45
It is the "tip" of the cone, and I found it easiest to path connect everything to. You could also connect any two points by moving around the cone, though writing out the equation is slightly harder (but still very do-able). – Potato Jun 2 '12 at 19:10
Thank yoy very much. I understand now. – srijan Jun 2 '12 at 20:01

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.

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Thanks i understand now. But i am finding it difficult to show mathematically path connectedness of subspace P. Though i am reading all the answers and trying to understand. – srijan Jun 2 '12 at 19:03