Example of Topological Group I'm trying to learn about topological groups but I can't seem to google anything that provides a simple and clear example of how the set of elements of a group correspond to open sets of a topology.
Also, what does it mean for a group binary operation to be continuous.  I would think that is always true since if $g_1\in G$ is open then then the inverse operation from $g_2\in G$ to $g_1$ would always map to an open discrete set.
 A: It might be easier to think of a topological group as a topological space with a group structure rather than a group with a topology.
Let $X$ be a topological space. This means that certain subsets of $X$ are in a family $\tau\subset\mathcal{P}(X)$ such that the usual axioms apply.
Now assuming the axiom of choice, there is a group structure on $X$. That is maps $m:X\times X\rightarrow X$ and $\operatorname{inv}:X\rightarrow X$ --- and an element $e\in X$ --- such that $(X,m)$ is a group.
When $m$ and $\operatorname{inv}$ are continuous then we say that $(X,m)$ is a topological group. 
This means that if $U\in\tau$, $m^{-1}(U)$ is an open set in the product topology of $X\times X$; and $\operatorname{inv}^{-1}(U)\in\tau$ also.
A: For your second paragraph, it might be helpful to see an example of a topological space with a discontinuous group operation.
The topological space is $X = \mathbb{R}$. I will pick a discontinuous bijection of $\mathbb{R}$: $f(x) = x$ if $x \ne 0,1$, $f(0)=1$, $f(1)=0$. Using this bijection, I will define a group operation $\oplus$ on $X$, having identity element $1$, using the formula 
$$x \oplus y = f^{-1}(f(x) + f(y))
$$
If $x_n=-1/n$ and $y_n=1/n$ then $\lim (x_n + y_n) = \lim 0 = 0$, but $\lim x_n + \lim y_n = 0 + 0 = 2$.
A: Perhaps try thinking about matrices. You know from linear algebra that the set of invertible $n \times n$ matrices is a group with group operation matrix multiplication. If you write out the functions that express the entries of the product of two matrices in terms of the entries of the matrices you can see that they are continuous (since they involve just sums of products of real numbers). That makes the set of invertible matrices a topological group.
The $1 \times 1$ invertible matrices are just the reals without $0$. It's pretty clear that multiplication there is continuous.
