Thread: Containment of a Symmetric Neighborhood

1. 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 $V \cdot V \subseteq U$?

2. 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.

3. 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?

4. 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!

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