1. ## Pathwise Connected

Problem: Show that the set S={u=(x,y,z) in R^3 | 2x^2 + y^2 + z^2 = 1} is pathwise-connected.

I plotted the function in MATLAB and attached it. In a way, this looks very similar to the unit sphere in $R^3$, which is pathwise connected where the parametric path is
f(t) = costu+sintw

where w is a point in the sphere that is perpendicular to u and v = -u. However, I am having a hard time to find a parametric path for this set.

The book says to use two previous exercises 1.) The union of two pathwise-connected subsets of R^n is pathwise-connected and 2.)the graph of the mapping F:A -> R^m is continuous then if A is pathwise-connected, then the set
G = {(u,v) in R^(n+m)|u in A, v = F(u))} is pathwise-connected.

Image:

Any help is greatly appreciated. Thank you.

2. Hello
Originally Posted by Paperwings
Problem: Show that the set S={u=(x,y,z) in R^3 | 2x^2 + y^2 + z^2 = 1} is pathwise-connected. (...)

The book says to use two previous exercises 1.) The union of two pathwise-connected subsets of R^n is pathwise-connected and 2.)the graph of the mapping F:A -> R^m is continuous then if A is pathwise-connected, then the set
G = {(u,v) in R^(n+m)|u in A, v = F(u))} is pathwise-connected.
I suggest you use the two previous exercises. Let . Let A be the domain of this two functions. (A is a subset of R^2) The set S can be described as the union of two sets :

Using , one can write

Is A pathwise connected ? Are F+ and F- continuous ?

3. Is A pathwise connected ? Are F+ and F- continuous ?

1. A is pathwise connected since there exists a

2. F+ and F- are continuous. Let , then the polynomial is continuous since polynomials are continuous.
Then let be the boundary of the the region.

For such that
then . Thus, F+ is continuous.
Similarly, F- is continuous.

Odd... latex is not working so I attached images.

4. Originally Posted by Paperwings
Is A pathwise connected ? Are F+ and F- continuous ?

1. A is pathwise connected since there exists a
If gamma is supposed to be a path between two points, why does it exist ?

I think it's easier to show that A={(x,y) in R^2 | 2x^2+y^2 <= 1} is pathwise connected because it is homeomorphic to {(x,y) in R^2 | x^2+y^2 <= 1} which is pathwise connected. This can be done using f(x,y)=(x/sqrt(2),y) for example.
2. F+ and F- are continuous. Let , then the polynomial is continuous since polynomials are continuous.
Then let be the boundary of the the region.

For such that
then . Thus, F+ is continuous.
Instead of using the epsilon-delta definition of continuity, you can simply say that g(x)=sqrt(x) is continuous on [0,infinity[ hence the composite function F+ is continuous on A.

Using these results you can conclude and say that S is pathwise connected.

5. Thank you for help. I haven't learned what a homeomorphic is nor is it in my book, but I'll look it up.