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Determine all of the open sets in given product topology

  1. Feb 18, 2017 #1
    1. The problem statement, all variables and given/known data

    ##X = \{1,2,3\}## , ##\sigma = \big\{\emptyset , \{1,2\}, \{1,2,3\} \big\}##, topology ##\{X, \sigma\}##
    ##Y = \{4,5\}## , ##\tau = \big\{\emptyset , \{4\}, \{4,5\} \big\}##, topology ##\{Y, \tau\}##
    ##Z = \{2,3\} \subset X##

    Find all the open sets in the subspace topology on ##Z## and determine all the open sets in the product topology on ##Z \times Y##

    2. Relevant equations


    3. The attempt at a solution

    As with previous threads, I'm hoping I'm on the right track. Just looking to check my answers.

    Subspace topology on ##Z##, ##\sigma_z = \big\{\emptyset , \{2\}, \{2, 3\} \big\}##

    Product topology is generated by basis ##\{ U \times V : U \in \sigma_z , V \in \tau \}##

    Thus the product topology on ##Z \times Y## should be created from basis:

    ##\beta = \big\{\emptyset, \{(2,4)\}, \{(2,4),(2,5)\}, \{(2,4),(2,5),(3,4),(3,5)\}, \{(4,2)\}, \{(4,2),(4,3)\}, \{(4,2),(5,2)\}, \{(4,2),(4,3),(5,2),(5,3)\}\big\} ##

    Thus all the open sets in the product topology on ##Z \times Y## are found in the union of all sets of basis elements in ##\beta## , plus the empty set:

    ## \Big\{\big( \cup U_\lambda \big)## ## \forall## ## U_\lambda \subset \beta\Big\} \cup \{\emptyset\}##

    Is this all okay? Thanks.

    (Many topology threads incoming - this subject is definitely my kryptonite)
     
  2. jcsd
  3. Feb 18, 2017 #2

    andrewkirk

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    The subspace topology of ##Z## has a basis with two non-empty sets, and the topology of ##Y## has a basis with two non-empty sets. So the product basis must have ##2\times 2=4## sets. The above list for ##\beta## has seven elements, so cannot be correct. Notice also that the first component of an element of the product set comes from ##Z##, which does not contain ##4##, yet the above list includes open sets containing elements whose first component is ##4##, so they cannot be from ##Z\times Y##.

    To get the basis of the product topology, just write out the four elements of the Cartesian product of the two bases.
     
  4. Feb 18, 2017 #3
    Ah ok, I think I see where I went wrong. For reasons unknown to me, I took the products ##Z \times Y## and ##Y \times Z##

    Second attempt, I'd go for:

    ##\beta = \big\{ \{(2,4)\}, \{(2,4),(2,5)\}, \{(2,4), (3,4)\}, \{(2,4),(2,5),(3,4),(3,5)\}\big\}##

    And then the open sets in that product topology are all possible unions of all possible subsets of ##\beta##, along with the empty set.
     
  5. Feb 18, 2017 #4

    andrewkirk

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    Yes, that looks right!
     
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