February 13th
Today I learned some basic results in topological group theory. What's interesting here is that topology on its own has the potential to be quite disgusting, but adding in a little group theory makes the condition significantly nicer. I think the high-level reason why is that topological grops put all elements on a more equal footing because we require, for any element $g,$\[U\text{ open}\iff gU\text{ open}\iff Ug\text{ open}.\]This condition is called homogeneity. In particular, knowing all (open) neighborhoods around $g$ is equivalent to knowing the (open) neighborhoods around $e$ and then multiplying by $g.$
As an example of this in effect, we show that $T_1$ (points are closed) implies $T_2$ (Hausdorff). As a preliminary, we remark that if we an open neighborhood $U$ of $e,$ then $U\cap U^{-1}$ is a symmetric open neighborhood containing $U$ of $e.$ Additionally, the group operation defines a map\[U\times U\to G.\]This has image containing $(e,e),$ so the image has nontrivial intersection with $U.$ Tracking the pre-image of $U$ will give an open neighborhood $V\subseteq U$ of $e$ with $VV\subseteq U.$ Combining, we can also assume that $V$ is symmetric. In essence, this lemma lets us multiply and invert elements "locally'' in a group with comfort.
Ok, now take $G$ a topological group which is $T_1.$ That is, for every pair of points $g,h\in G,$ there exists an open neighborhood of $g$ excluding $h.$ We show that $G$ is $T_2,$ which means for every pair $g,h\in G,$ there exist disjoint open neighborhoods of $g$ and $h.$
Using the intuition from before, we focus on the identity case where, say $h=e.$ Using the $T_1$ condition, we fix a neighborhood $U$ around $h=e$ excluding $g.$ We would like to use $gU$ for our neighborhood around $g,$ but $U$ and $gU$ might not be disjoint. However, any intersection\[u_2=gu_1\in U\cap gU\]forces $g=u_2u_1^{-1}$ to almost live inside of $U.$ So to fix this, we use the lemma to take a smaller open neighborhood $V\subseteq U$ of $e$ to be symmetric with $VV\subseteq U.$ Then we use $V$ and $gV$ as our neighborhoods, giving $u_1^{-1}\in V,$ and $g=u_2u_1^{-1}\in VV\subseteq U.$ This is our contradiction, proving $T_2.$
For clarity, let's do the general case with $h\ne e$ more directly. By $T_1,$ fix an open neighborhood $U$ around $e$ excluding $h^{-1}g,$ and the lemma gives us a smaller open neighborhood $V\subseteq U$ of $e$ which is symmetric and satisfies $VV\subseteq U.$ Now we claim\[gV\quad\text{and}\quad hV\]are the required neighborhoods for $T_2.$ Indeed, if $gV\cap hV$ is non-empty, then $h^{-1}gV\cap V$ is non-empty, so\[\{h^{-1}g\}\cap VV^{-1}\]is also non-empty. However, $V^{-1}=V,$ so $VV^{-1}\subseteq VV\subseteq U,$ and $h^{-1}g\notin U,$ so this is a contradiction. Thus, $gV\cap hV=\emp,$ and we are done here.