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

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

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

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

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

2. Originally Posted by huram2215
This is from Munkres, 2nd ed., Ch2, supp. exercise 7(a).

Definitions:

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

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

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

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

I can show that if V is symmetric, that $\displaystyle V \cdot V$ is an open set in $\displaystyle G$ and $\displaystyle e \in V \cdot V$ trivially. But I'm not seeing how to prove that a construction of a symmetric V from U has $\displaystyle V \cdot V \subseteq U$?
The continuity of the multiplication map $\displaystyle \mu$ means that there exist neighbourhoods $\displaystyle V,\, W$ of $\displaystyle e$ such that $\displaystyle \mu(g_1,g_2)\in U$ for all $\displaystyle g_1\in V$ and $\displaystyle g_2\in W$. But that is the same as saying that $\displaystyle V\cdot W \subseteq U$. Replacing $\displaystyle V$ and $\displaystyle W$ by $\displaystyle V\cap W$, we may assume that $\displaystyle W=V$, so that $\displaystyle V\cdot V\subseteq U$. Finally, replacing $\displaystyle V$ by $\displaystyle V\cap V^{-1}$, we may assume that $\displaystyle V$ 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 $\displaystyle V$ and $\displaystyle W$. Given $\displaystyle U$ open in $\displaystyle G$, since $\displaystyle \mu$ is continuous we know that $\displaystyle \mu^{-1}(U)$ is open in $\displaystyle G \times G$ which can be represented as the union of basis elements from G as $\displaystyle \mu^{-1}(U) = \cup_{\lambda \in \Lambda}{(V_\lambda \times W_\lambda)}$. The union doesn't pull through the product, so this set does not factor nicely into separate $\displaystyle V$ and $\displaystyle W$.

Example: the open unit ball in $\displaystyle \mathbb{R}^2$ 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 $\displaystyle V$ and $\displaystyle W$. Given $\displaystyle U$ open in $\displaystyle G$, since $\displaystyle \mu$ is continuous we know that $\displaystyle \mu^{-1}(U)$ is open in $\displaystyle G \times G$ which can be represented as the union of basis elements from G as $\displaystyle \mu^{-1}(U) = \cup_{\lambda \in \Lambda}{(V_\lambda \times W_\lambda)}$. The union doesn't pull through the product, so this set does not factor nicely into separate $\displaystyle V$ and $\displaystyle W$.

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

Example: the open unit ball in $\displaystyle \mathbb{R}^2$ contains the product $\displaystyle (-\frac12,\frac12)\times(-\frac12,\frac12)$ of two open intervals!

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