[Math] Topological Group, symmetric neighborhood, Hausdorff, disjoint open sets

general-topologytopological-groups

Let $G$ be a topological group with identity $e$. If $A, B$ are subsets of $G$, we let $A * B$ denote the collection of elements $a * b$ for $a \in A, b \in B$, and we let $A^{-1}$ denote the set of elements $a^{-1}$ for $a \in A$.

a. A neighborhood $V$ of $e$ is said to be symmetric if $V^{-1} = V$. Show that if $W$ is any neighborhood of $e$, then there exists a symmetric neighborhood $V$ of $e$ such that $V * V \subset W$.

b. Suppose that $G$ is $T_1$ as a topological space. Show that then $G$ is necessarily Hausdorff. Specifically, if $x \neq y$ are given, show that there exists a neighborhood $V$ of the identity such that $V * x$ and $V * y$ are disjoint.

c. Show that if $G$ is $T_1$, it is in fact regular. Recall that a space $X$ is said to be regular if for all $x \in X$ and all closed subsets $A \subset X$ not containing $x$, there exist disjoint open sets $V, W$ such that $x \in V, A \subset W$.

Best Answer

Hint:

a. For an arbitrary nbhood $W$ of $e$, let $U=W\cap W^{-1}$, then plainly $U=U^{-1}$, $U$ is a nbhood of $e$, and $U\subset W$. (now use, theorem: Neighborhood Axioms of $e$).

b. If $G$ is a topologial group the following conditions are equivalent:

i) $G$ is a $T_0$ space.

ii) $G$ is a $T_1$ space.

iii) $G$ is a $T_2$ space.

iv) If $\beta_e$ is a fundamental system of neighborhoods of $e$ then $\bigcap β e ={e}$.

v) $\{e\}$ is a closed subgroup of $G$.

c. Every topological group is regular.

I'll refer you to a few books on topological groups. Hopefully useful to you.

  1. A. Arhangel'skii, M. Tkachenko, Topological Groups and Related Structures.

  2. T. Husain, Introduction to topological groups.

  3. E. Hewitt, K. A. Ross, Abstract Harmonic Analysis: Volume 1.