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Product Spaces

  1. Oct 27, 2005 #1
    If [itex]X[/itex] and [itex]Y[/itex] are two connected topological spaces then so is [itex]X \otimes Y[/itex].

    I want to understand the proof of this theorem but I am having some difficulties. Even though we went over it in class, it is still unclear to me.

    The professor constructed this continuous function:

    [tex]f:X\otimes Y \rightarrow \{0,1\}[/tex]

    Where [itex]\{0,1\}[/itex] is a discrete topological space. Then he shows that [itex]f[/itex] is constant. He then claimed that [itex]\{x\} \otimes Y[/itex] is homeomorphic with [itex]Y[/itex] hence this subspace ([itex]\{x\}\otimes Y[/itex]) is connected - since [itex]Y[/itex] is.

    Now this does not make sense to me and I wouldn't be suprised if it didn't make sense to any of you. If you think you have a better way of explaining the proof (it doesn't have to be this one) then I would appreciate the effort.

    Im not sure exactly why one would begin by setting up a continuous function which maps points in the product space [itex]X \otimes Y[/itex] to either 0 or 1 in the discrete topological space, then claim that the function constant - then go on to show that [itex]\{x\} \otimes Y[/itex] is homeomorphic to [itex]Y[/itex].??
  2. jcsd
  3. Oct 27, 2005 #2
    Actually come to think of it, I do understand this.

    Im just not sure of why this is the correct procedure to proving that the product space is connected.
  4. Oct 27, 2005 #3
    Hmmm, perhaps it is because if one finds a concrete homeomorphism between the product space to a pre-determined connected topological space, then since the homeomorphism preserved topological structure, the product space is therefore connected? Could this be the reason?
  5. Oct 27, 2005 #4


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    I think you may have misunderstood what he was saying. Of course, [itex]\{x\} \times Y[/itex] is homeomorphic to Y: the function f(x,y)= y is clearly continuous with a continuous inverse f-1(y)= (x,y).
    That being true, since Y is connected, so is [itex]\{x\} \times Y[/itex]. A continuous function maps connected sets to connected sets so any continuous function must map [itex]\{x\} \times Y[/itex] to either {0} or {1}- those are the only connected subsets of {0, 1} with the discreet topology.
    Now, do same with [itex]X \times \{y\}[/itex]- and observe that any y will be matched with every x, any x with every y: f must map all of [itex]X \times Y[/itex] into a connected set: either {0} or {1} and therefore f is a constant.
    Your professor is not first proving that f must be constant and then that f map [itex]X \times Y[/itex] to a connected set, he is first proving that f maps [itex]X\times Y[/itex] to a connected set and then using that to prove that f must be constant.
    Last edited: Oct 27, 2005
  6. Oct 27, 2005 #5
    Thanks Halls once again. I understand it now.

    However, one small question. Why choose {0,1} as the target space for f? I need motivation for the choice of {0,1}. Why not X or Y or some other space? Is it because {0,1} is easy to work with? Im not sure.
  7. Oct 27, 2005 #6
    The latter reason. He wanted a disconnected space to map to where it was easy to characterize when the image of another space was disconnected.
  8. Oct 27, 2005 #7
    Perfect. Exactly what I thought (I couldn't be sure). Thanks hyper.
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