Is Path-Connectedness Provable by Arbitrary Points in a Metric Space?

  • Thread starter Thread starter jessicaw
  • Start date Start date
  • Tags Tags
    Path
jessicaw
Messages
54
Reaction score
0

Homework Statement


For any S in X, show that S is path-connected if and only if there exists p in S such that any point x in S can be joined to p by a path.

Homework Equations


A metric space is path connected if any 2 points can be joined by a path in that metric space.

The Attempt at a Solution


If part: Well..if there exists such p, let a joined to p and b joined to p, then a can be joined to b, so arbitary a,b can be joined?
Only if part: Now any 2 points can be joined by a path in that metric space, so let p be a fixed point lying on the path of ab, so p can be joined to a and can be joined to b, so p can be joined to every point?The above is an attempt but there is some errors in the proof, can you check my attempt? Also i think this question is not so easy(i believe the proof should require use of advanced stuff like continuous function, 2 valued, [0,1],...and the like), right?
 
Physics news on Phys.org
The if-part seems correct.
For the only if part, you take p lying on the path from a to b, and then you show a and b can be joined to p. But you have to show that EVERY point can be joined to p. You've only shown it for a and b. Or did I miss something?
 
micromass said:
The if-part seems correct.
For the only if part, you take p lying on the path from a to b, and then you show a and b can be joined to p. But you have to show that EVERY point can be joined to p. You've only shown it for a and b. Or did I miss something?

Yes i am concerned about this. My argument is a certain path connecting arbitray a,b. can be created so that p, being a fixed point, lies on this path connecting arbitray a,b.
 
But won't it be easier to take p arbitrary. And then showing that every point can be connected to p? Just use path-connectedness...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top