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

In summary, the intermediate value property states that if a point lies between two points in an ordered set, then that point has a value equal to the distance between the two points.
  • #1
GridironCPJ
44
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.
 
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  • #2
be more precise. give definitions and full statements and we will have a better chance of answering.
 
  • #3
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.
 
  • #4
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.
 
  • #5
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.
 
  • #6
Oh, but that is even easier. Take a seperation, map the two open sets to different points.
 
  • #7
micromass said:
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.
 
  • #8
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?
 
  • #9
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.
 

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

1. How does the intermediate value property relate to connectedness?

The intermediate value property states that if a function is continuous on a closed interval [a, b], then it takes on every value between f(a) and f(b). This means that the function's graph must be connected, which is a key component of connectedness.

2. Can you give an example of a function that satisfies the intermediate value property and is also connected?

One example is f(x) = x^2, defined on the interval [-1, 1]. This function is continuous on the closed interval and takes on every value between f(-1) = 1 and f(1) = 1, making it satisfy the intermediate value property. Its graph is also a parabola, which is a single connected curve.

3. How would you prove that the intermediate value property implies connectedness?

To prove this, we can use a proof by contradiction. Assume that the intermediate value property holds for a function f on a closed interval [a, b], but the graph of f is not connected. This means that there exists two points c and d such that f(c) < f(d) and there exists a point e between f(c) and f(d) that is not in the range of f. However, this contradicts the intermediate value property, so our assumption must be false and the function must be connected.

4. Is the intermediate value property a sufficient condition for connectedness?

No, the intermediate value property is not a sufficient condition for connectedness. It only guarantees that the graph of a function is connected, but there are other requirements for a set to be considered connected, such as being a single, unbroken curve.

5. Can you provide a real-world application of the intermediate value property and connectedness?

The intermediate value property and connectedness are often used in economics to analyze supply and demand curves. The intermediate value property ensures that there is a price at which the quantity supplied and the quantity demanded are equal, which is known as the market equilibrium. The connectedness of the curves ensures that there is a continuous relationship between price and quantity, allowing for accurate analysis and prediction of market behavior.

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