To unite topology and group theory, we can define a topological group as a group $(G, \ast)$ with a topology such that $(x, y) \mapsto x \ast y$ and $x \mapsto x\inv$ are continuous functions. Only requiring the first to be continuous is not enough.

Some sources like Munkres additionally requires it to be $T_1$, which seemed arbitrary at first, why not Hausdorff instead? It turns out that (1) being Hausdorff (2) being $T_1$ (3) $\{e\}$ being a closed set; are all equivalent statements for topological groups. So one interesting way to prove some space is Hausdorff is to define a group operation on the space which is continuous and its inverse is continuous too, then prove that $\{e\}$ is closed. This is probably useless.

Theorem 1: Take $\R$ with the topology of open rays $$\tau = \set{ (-\infty, a) \given a \in \R }$$ Then for the group $(\R, +)$, $(x, y) \mapsto x + y$ is continuous, but $x \mapsto -x$ isn't.

Proof. Since every open set is of the form $(-\infty, a)$, we just prove every element $(x, y) \in +\inv((-\infty, a))$ is an interior element. If $x + y < a$, then we can find some $\delta > 0$ such that $x + y + 2\delta < a$. We define the open set $$(-\infty, x + \delta) \times (-\infty, y + \delta)$$ If $(x', y')$ is in this set, then $x' < x + \delta$ and $y' < y + \delta$, which means $$x' + y' < x + y + 2\delta < a$$ so $(x, y) \subseteq (-\infty, x + \delta) \times (-\infty, y + \delta) \subseteq +\inv((-\infty, a))$ as we wanted.

Negation isn't continuous, however, since the inverse image of $(-\infty, 0)$ would be $(0, \infty)$. This set is not bounded above, so it isn't open under the topology we defined. $\square$

Theorem 2: If $(G, \ast)$ is a topological group, then $(x, y) \mapsto x \ast y\inv$ is continuous.

Proof. The identity function $\text{id}$ is continuous, and the inverse function (we call it $\text{inv}$) is continuous. The product of two continuous functions is continuous. $\ast$ is continuous. Therefore, the composition $$\ast \circ (\text{id} \times \text{inv})$$ is continuous. $\square$

Theorem 3: If $(G, \ast)$ is a topological group, then the following three statements about its topology are equivalent: (1) it is Hausdorff (2) it is $T_1$ (3) $\{e\}$ is closed.

Proof. $(1) \implies (2) \implies (3)$ is obvious. We prove $(3) \implies (1)$. Suppose $\{e\}$ is closed. Using theorem 2, if we define $f(x, y) = x \ast y\inv$, then it is continuous and $K = f\inv(\{e\})$ is a closed set. But $(x, y) \in K$ if and only if $x \ast y\inv = e$ which can only mean $x = y$. Therefore, $K = \Delta = \set{ (x, x) \given x \in G }$ is closed, and so the space is Hausdorff. $\square$