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Math Help - Containment of a Symmetric Neighborhood

  1. #1
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    Containment of a Symmetric Neighborhood

    This is from Munkres, 2nd ed., Ch2, supp. exercise 7(a).

    Definitions:

    Topological Group: ((G , \cdot), \tau_G) is a topological group with identity e. This means that the product function \mu (g_1,g_2) = g_1 \cdot g_2 is continuous and I(g_1) = g_1^{-1} is a continuous map.

    For A, B subsets of G, A \cdot B = \{ a \cdot b : a \in A, b \in B\}. A^{-1} = \{ a^{-1} : a \in A \}.

    Symmetric neighborhood: A neighborhood V of e such that V = V^{-1}. Note that, for U open in G and e \in U, U \cap U^{-1} and U \cdot U^{-1} are both symmetric.

    Claim: Any neighborhood U of e contains a symmetric neighborhood V \subseteq U such that V \cdot V \subseteq U.

    I can show that if V is symmetric, that V \cdot V is an open set in G and e \in V \cdot V trivially. But I'm not seeing how to prove that a construction of a symmetric V from U has V \cdot V \subseteq U?

    Thank you for your help.
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  2. #2
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    Quote Originally Posted by huram2215 View Post
    This is from Munkres, 2nd ed., Ch2, supp. exercise 7(a).

    Definitions:

    Topological Group: ((G , \cdot), \tau_G) is a topological group with identity e. This means that the product function \mu (g_1,g_2) = g_1 \cdot g_2 is continuous and I(g_1) = g_1^{-1} is a continuous map.

    For A, B subsets of G, A \cdot B = \{ a \cdot b : a \in A, b \in B\}. A^{-1} = \{ a^{-1} : a \in A \}.

    Symmetric neighborhood: A neighborhood V of e such that V = V^{-1}. Note that, for U open in G and e \in U, U \cap U^{-1} and U \cdot U^{-1} are both symmetric.

    Claim: Any neighborhood U of e contains a symmetric neighborhood V \subseteq U such that V \cdot V \subseteq U.

    I can show that if V is symmetric, that V \cdot V is an open set in G and e \in V \cdot V trivially. But I'm not seeing how to prove that a construction of a symmetric V from U has V \cdot V \subseteq U?
    The continuity of the multiplication map \mu means that there exist neighbourhoods V,\, W of e such that \mu(g_1,g_2)\in U for all g_1\in V and g_2\in W. But that is the same as saying that V\cdot W \subseteq U. Replacing  V and W by V\cap W, we may assume that W=V, so that V\cdot V\subseteq U. Finally, replacing  V by V\cap V^{-1}, we may assume that  V is symmetric.
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  3. #3
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    I think there is still a gap.

    Your result looks close, but there is one part that confuses me still; the existence of V and W. Given U open in G, since \mu is continuous we know that \mu^{-1}(U) is open in G \times G which can be represented as the union of basis elements from G as \mu^{-1}(U) = \cup_{\lambda \in \Lambda}{(V_\lambda \times W_\lambda)}. The union doesn't pull through the product, so this set does not factor nicely into separate V and W.

    Example: the open unit ball in \mathbb{R}^2 is not the product of two open intervals. Am I missing something?
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  4. #4
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    Quote Originally Posted by huram2215 View Post
    Your result looks close, but there is one part that confuses me still; the existence of V and W. Given U open in G, since \mu is continuous we know that \mu^{-1}(U) is open in G \times G which can be represented as the union of basis elements from G as \mu^{-1}(U) = \cup_{\lambda \in \Lambda}{(V_\lambda \times W_\lambda)}. The union doesn't pull through the product, so this set does not factor nicely into separate V and W.

    Example: the open unit ball in \mathbb{R}^2 is not the product of two open intervals. Am I missing something?
    But product sets form a basis for the product topology, so \mu^{-1}(U) will contain a neighbourhood of e\times e of the form V\times W.

    Example: the open unit ball in \mathbb{R}^2 contains the product (-\frac12,\frac12)\times(-\frac12,\frac12) of two open intervals!
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  5. #5
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    Ah, yes. Of course - I should have thought of that. Thanks again for your help.
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