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

  1. May 16, 2012 #1
    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.
  2. jcsd
  3. May 16, 2012 #2


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    be more precise. give definitions and full statements and we will have a better chance of answering.
  4. May 17, 2012 #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.
  5. May 17, 2012 #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.
  6. May 17, 2012 #5
    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.
  7. May 17, 2012 #6
    Oh, but that is even easier. Take a seperation, map the two open sets to different points.
  8. May 17, 2012 #7
    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.
  9. May 17, 2012 #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?
  10. May 26, 2012 #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
  11. May 26, 2012 #10


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    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?
  12. May 26, 2012 #11
    i just did in my head
  13. May 27, 2012 #12


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