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

[GridironCPJ]

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.

 

[mathwonk]

be more precise. give definitions and full statements and we will have a better chance of answering.

 

[GridironCPJ]

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.

 

[micromass]

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.

 

[GridironCPJ]

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.

 

[micromass]

Oh, but that is even easier. Take a seperation, map the two open sets to different points.

 

[GridironCPJ]

Oh, but that is even easier. Take a seperation, 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.

 

[GridironCPJ]

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?

 

[algebrat]

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

 

[mathwonk]

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?

 

[algebrat]

i just did in my head

 

[mathwonk]

i think he meant, if the implication holds for all ordered Y, then X is connected.