> [!definition] > > Let $G$ be a [[Group|group]] and $\topo$ be the collection of [[Open Set|open sets]] over $G$. The [[Topological Space|topology]] $(G, \topo)$ forms a **topological group** if > $ > (x, y) \mapsto xy \quad x \mapsto x^{-1} > $ > the above maps are [[Continuity|continuous]]. > [!theorem] > > Let $(G, \topo)$ be a topological group, $x \in G$, $U \in \topo$. Then > - $xU, Ux \in \topo$. > - $U^{-1} \in \topo$. > > *Proof*. $x^{-1}xU = U$, so the preimage of $U$ by multiplying $x^{-1}$ is $xU$, which makes it open (?). > > $U$ is the preimage of $U^{-1}$ by the inverse map, so $U^{-1}$ is also open by continuity. > [!theorem] > > Let $U$ be a [[Neighbourhood|neighbourhood]] of $e \in G$, then there exists a symmetric neighbourhood $V$ of $e$ such that $VV \subseteq U$ where $VV = \bracs{v_1v_2: v_1, v_2 \in V}$. > > *Proof*. Since $U$ is a neighbourhood of $e \in G$, the preimage of $\inte(U)$ under the composition map is an open set containing $(e, e)$. The preimage of $U$ would be a neighbourhood of $(e, e)$. > > Since the preimage of $U$ is a neighbourhood, > $ > \exists W \subseteq X: W \times W \subseteq M^{-1}(U) \Rightarrow WW \subseteq U > $ > where $W$ is a neighbourhood of $e$. Now, take $V = W \cap W^{-1}$, which is still a neighbourhood of $e$, then $V$ has the desired properties.