How would you show that the intermediate value property implies connectedness?

  • Context: Graduate 
  • Thread starter Thread starter GridironCPJ
  • Start date Start date
  • Tags Tags
    Property Value
Click For Summary

Discussion Overview

The discussion revolves around the relationship between the intermediate value property (IVP) and the connectedness of a space X. Participants explore how to demonstrate that if a space has the IVP, it must be connected, and they consider various proof strategies, including contradiction.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that if a space X has the IVP, then it is connected, suggesting a proof by contradiction.
  • Another participant emphasizes the need for precise definitions and full statements to facilitate the discussion.
  • A definition of the IVP is provided, along with a description of connectedness in terms of separations.
  • Several participants suggest using contradiction, assuming the absence of the IVP and attempting to find disjoint open sets in Y whose preimage covers all of X.
  • One participant suggests mapping two open sets from a separation of X to different points in Y to demonstrate a contradiction.
  • Concerns are raised about the assumption that the images of disjoint sets must also be disjoint, with one participant questioning the implications of continuity in this context.
  • A later reply provides a justification for the proposed strategy, discussing the implications of continuity and the order topology on the separation of X.
  • Another participant presents a counterexample where a space satisfies the IVP but is not connected, challenging the earlier claims and suggesting a flaw in the reasoning.
  • Further discussion includes a reference to the relationship between connectedness and continuous maps to a two-point set.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the IVP for connectedness, with some supporting the idea while others present counterexamples and challenge the assumptions made in the proofs. The discussion remains unresolved with multiple competing perspectives.

Contextual Notes

There are limitations in the assumptions made regarding the images of disjoint sets and the implications of continuity. The discussion also highlights the need for precise definitions and the potential for counterexamples that challenge the proposed proofs.

GridironCPJ
Messages
44
Reaction score
0
Suppose a space X has the intermediate value property (f: X->Y continuous, Y has the order topology), then X is connected.

How would you show this? This is just the converse of the intermediate value property.
 
Physics news on Phys.org
be more precise. give definitions and full statements and we will have a better chance of answering.
 
I will define the intermediate value property/theorem exactly as it is expressed in Munkres.

(Intermediate vaue theorem) Let f: X->Y be a continuous map, where X is a connected space and Y is an ordered set in the order topology. If a and b are two points of X and if r is a point of Y lying between f(a) and f(b), then there exists a point c of x such that f(c)=r.

Connectedness of X: There does not exist a separation of X (a separation would be a pair of disjoint nonempty subsets whose union is in X).

Potential gameplan for proof: Contradiction seems to fit comfortably in proofs involving connectedness. I havn't been able to come up with anything worthwhile.
 
Contradiction is the way to go.
Assume that we don't have the intermediate value property. So a point r is not in the image. Find two disjoint open sets in Y whose preimage is entire X.

It might help you to find a disconnected space and see why the intermediate value property fails.
 
micromass said:
Contradiction is the way to go.
Assume that we don't have the intermediate value property. So a point r is not in the image. Find two disjoint open sets in Y whose preimage is entire X.

It might help you to find a disconnected space and see why the intermediate value property fails.

That proves the converse of what I'm trying to prove (connectedness of X => IVP). We assume X is connected, then we state that X does not have the IVP, split Y into 2 disjoint nonempty open sets, since f is continuous, then the disjoint open sets have a preimage that is disjoint and open in X, thus X is not connected, contradition.

I'm trying to figure out how you prove (IVP => connectedness of X). I've attempted contradiction, but I get nowhere.
 
Oh, but that is even easier. Take a separation, map the two open sets to different points.
 
micromass said:
Oh, but that is even easier. Take a separation, map the two open sets to different points.

I was thinking of something similar, where we assume the IVP, then suppose X is separated into A and B, by which the two disjoint, nonempty open sets map into Y s.t. the images are disjoint and their closures do not intersect, that way, there are intermediate points between f(A) and f(B) that do not have a preimage. The only problem I had with this idea though is that it's assuming that the image of A and B are disjoint, but this is not necessarily true. What if f(A)=f(B)? All we have is continuity of f, so I'm guessing some sort of restriction to the domain or the codomain (or both) might be necessary. I'm not entirely sure though.
 
I think I've got some good justification for the strategy you mentioned:

A and B are a separation of X. Since f is continuous, there must exist C and D in Y such that f^-1(C)=A and f^-1(D)=B. Also, C and D cannot interest, for if they do, then their intersection elements are mapped into both A and B, which cannot be. Also, since Y is in the order topology, the open sets in Y are open intervals so there are points "in between" two disjoint open intervals. We can let f(a) be in C and f(b) be in D, then there is an r in between C and D s.t. there does not exist an f(c)=r since r is in neither C or D, which are the only open sets that map into X by f^-1. Thus, the IVP is not satisfied, a contradiction.

Does this seem accurate?
 
i have a counter example. Let X be the union of 1 and 2. let f map both elements to 3. then f satisfies the IVP (i don't know why you say X satisfies IVP). but X is not connected. if you keep this simple counterexample in mind, you should be able to find the error in your proof above, for example, showing C and D do not intersect, that contradicts my counterexample.

i think you have the problem statement written incorrectly
 
  • #10
have you ever checked that a space X is connected iff every continuous map from X to the two point set {0,1} is constant?
 
  • #11
i just did in my head
 
  • #12
i think he meant, if the implication holds for all ordered Y, then X is connected.
 

Similar threads

Undergrad Does topology distinguish between real and imaginary dimensions?
  • · Replies 35 ·
2
Replies
35
Views
5K
Undergrad Homemorphism in quotient topology
  • · Replies 5 ·
Replies
5
Views
2K
MHB Intermediate Value Theorem ....Silva, Theorem 4.2.1 .... ....
  • · Replies 3 ·
Replies
3
Views
3K
Undergrad Convergence not defined by any metric
  • · Replies 17 ·
Replies
17
Views
2K
Graduate Question about existence of path-lifting property
  • · Replies 8 ·
Replies
8
Views
3K
Undergrad How Do Metric Space Properties Relate to Continuity?
  • · Replies 8 ·
Replies
8
Views
2K
Graduate Question about lifting of a function
  • · Replies 2 ·
Replies
2
Views
3K
Undergrad A non-empty intersection of closures of level sets implies discontinuity
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Proving that convexity implies second order derivative being positive
  • · Replies 6 ·
Replies
6
Views
3K
MHB Show Existence of $a,b$: Proof for $f'(\xi) \neq 0$
  • · Replies 11 ·
Replies
11
Views
2K