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Open sets and closed sets in product topology

  1. Apr 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Let [tex](X_a, \tau_a), a \in A[/tex] be topological spaces, and let [tex]\displaystyle X = \prod_{a \in A} X_a[/tex].

    2. Relevant equations

    1. Prove that the projection maps [tex]p_a : X \to X_a[/tex] are open maps.

    2. Let [tex]S_a \subseteq X_a[/tex] and let [tex]\displaystyle S = \prod_{a \in A} S_a \subseteq \prod_{a \in A} X_a[/tex]. Prove that [tex]S[/tex] is closed iff each [tex]S_a \subseteq X_a[/tex] is closed.

    3. Let [tex]T_a \subseteq X_a[/tex], prove that [tex]\displaystyle \overline{\prod_{a \in A} T_a} = \prod_{a \in A} \overline{T_a}[/tex].

    4. If [tex]\abs{A} \leq \abs{\mathbb{N}}[/tex] and each [tex]X_a[/tex] is separable, prove that [tex]X[/tex] is separable.

    3. The attempt at a solution

    I don't know how to prove open/closed set problems in product topology. Can someone please give me some hint as to how I should approach these proofs? Some hints on each question will be even better.

    1. This means that any open subset of the product space [tex]X[/tex] remains open when projected down to the [tex]X_\alpha[/tex].

    Is it because the production topology [tex]\tau[/tex] for [tex]X[/tex] is the weakest topology with regard to [tex]\{ p_a :X \to X_a | a \in A \}[/tex]?

    2.

    3.

    4. [tex]X_a[/tex] is separable means there is a countable subset [tex]S_a \subseteq X_a[/tex] such that [tex]\overline{S_a} = X_a[/tex].
     
  2. jcsd
  3. Apr 25, 2010 #2
    Write out what an open set in the product topology looks like: http://en.wikipedia.org/wiki/Product_topology#Definition (the second paragraph).

    Based on this, where does the projection map send an arbitrary open set?

    For (4), use (3): take the product of the Sa and show that it's closure is X.
     
  4. Apr 26, 2010 #3
    Thank you very much for your reply. Here are my answers so far based on your suggestions.

    1. The product topology [tex]\tau[/tex] is generated from base [tex]\mathfrak{B}[/tex] consisting of product sets [tex]\displaystyle \prod_{a \in A} U_a[/tex] where only finitely many factors are not [tex]X_a[/tex] and the remaining factors are open sets in [tex]X_a[/tex]. Therefore the project [tex]p_a[/tex] projects an open set [tex]S \subseteq X[/tex] to either [tex]X_a[/tex] or some open subset [tex]S_a \subset X_a[/tex].

    2.

    3.

    4. [tex]X_a[/tex] is separable means there is a countable subset [tex]S_a \subseteq X_a[/tex] such that [tex]\overline{S_a} = X_a[/tex]. Using previous result, we have
    [tex]
    \begin{align*}
    \prod_{a \in A} \overline{S_a} = \prod_{a \in A} X_a = \overline{\prod_{a \in A} S_a} = X
    \end{align*}
    [/tex]
    Since [tex]S_a[/tex] is countable and [tex]|A| \leq |\mathbb{N}|[/tex], the cartesian product [tex]\displaystyle \prod_{a \in A} S_a[/tex] is countable. Hence [tex]X[/tex] is separable.

    ***This could be wrong, if [tex]|A| = |\mathbb{N}|[/tex], then the cartesian product does not have to be countable. So what is the set separable? Should the question say If [tex]|A| < |\mathbb{N}|[/tex] and each [tex]|X_a|[/tex] is separable, prove that [tex]|X|[/tex] is separable?
     
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