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

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Homework Help Overview

The discussion revolves around the concept of path-connectedness in metric spaces, specifically examining the conditions under which a set S is path-connected based on the existence of a point p within S that can connect to any other point in S via a path.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of having a fixed point p and whether it can connect all points in S. There is an attempt to clarify the proof structure, particularly distinguishing between the 'if' and 'only if' parts of the statement. Questions arise about the completeness of the argument regarding the connection of all points to p.

Discussion Status

Some participants express concerns about the adequacy of the original proof attempt, particularly in demonstrating that every point can be connected to p. Others suggest alternative approaches, such as considering an arbitrary point p and using the definition of path-connectedness to establish connections.

Contextual Notes

There is an acknowledgment of potential errors in the proof attempt, and participants are considering the need for advanced concepts such as continuous functions in their reasoning. The complexity of the proof is noted, with some participants suggesting that it may not be straightforward.

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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?
 
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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...
 

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