Containment of a Symmetric Neighborhood

**huram2215** · November 9th 2010, 09:00 AM

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.

**Opalg** · November 9th 2010, 11:12 AM

Originally Posted by huram2215

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.

**huram2215** · November 9th 2010, 12:38 PM

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?

**Opalg** · November 9th 2010, 01:03 PM

Originally Posted by huram2215

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!

**huram2215** · November 9th 2010, 04:10 PM

Ah, yes. Of course - I should have thought of that. Thanks again for your help.

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I think there is still a gap.

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