Why is the metric tensor symmetric? I was reading Schutz, A First Course in General Relativity. On page 9, he argued that the metric tensor is symmetric:

$$
ds^2~=~\sum_{\alpha,\beta}\eta_{\alpha\beta} ~dx^{\alpha}~dx^{\beta}
$$
  $\text{Note that we can suppose}$ $\eta_{\alpha\beta}=\eta_{\beta\alpha}$ $\text{for all}$ $\alpha$ $\text{and}$ $\beta$ $\text{since only the sum}$ $\eta_{\alpha\beta}+\eta_{\beta\alpha}$ $\text{ever appears in the above equation when}$ $\alpha\neq\beta$.

I don't understand his argument. If someone can explain why, I would really appreciate it.
 A: The metric tensor is created from the spacetime interval equation. On top of that, $[dx^\alpha,dx^\beta]=0$. Suppose we have a 1+1 dimensional spacetime, if you are given an interval equation resembling:
$$ds^2=-a\,dx_0^2+b\,dx_1^2+c\,dx_0dx_1$$
Obviously, $\eta_{0\,0}=-a$ and $\eta_{1\,1}=b$. Now, you can assume that, say, $\eta_{0\,1}=c/3$ and $\eta_{1\,0}=2c/3$ such that you still get the $c\,dx_0dx_1$ term, but why would you? You know $c=\eta_{1\,0}+\eta_{0\,1}$ and that's the only requirement you have. Since $dx_0$ and $dx_1$ commute, it is easier for everybody if you just say the tensor is symmetric and set $\eta_{1\,0}=\eta_{0\,1}=c/2$. Nothing changes except that you have saved yourself some trouble later on.
A: Assume $\eta_{\alpha\beta} \neq \eta_{\beta\alpha}$. Because it's irrelevant what letter we use for our indices, 
$$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}.$$ Then $$\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}$$ so only the symmetric part of $\eta_{\alpha\beta}$ would survive the sum. As such we may as well take $\eta_{\alpha\beta}$ to be symmetric in its definition. 
