determine basis for topology on $\mathbb{R}^2$ Determine whether the collection of subsets below form a basis for a topology on $\mathbb{R}^2$.
All subsets of the form $T_{\epsilon}(x)=\lbrace (y_1,y_2) : |x_1+x_2-y_1-y_2| <\epsilon \rbrace$ for all $x \in \mathbb{R}^2$ and all $\epsilon>0$
By doing some algebra, we obtain $-\epsilon -(x_1+x_2) < y_1+y_2<\epsilon -(x_1+x_2)$, which means that the set is the region bounded by two straight lines with negative gradient and y-intercept $-\epsilon -(x_1+x_2)$ and $\epsilon -(x_1+x_2)$ respectively.
The answer given is that the collection of the subsets above does not form a basis for topology on $\mathbb{R}^2$. Why? I thought every point is contained in one of the subsets above and also intersection between any two subsets is again the region bounded by two straight lines. I don't see why it fails to be a basis.
 A: You have a sign wrong: $T_\epsilon(x)$ is the set of $y=\langle y_1,y_2\rangle$ satisfying
$$(x_1+x_2)-\epsilon<y_1+y_2<(x_1+x_2)+\epsilon\;.$$
However, you’re right: $\{T_\epsilon(x):x\in\Bbb R^2\text{ and }\epsilon>0\}$ is a base for a topology $\tau$ on $\Bbb R^2$.
In fact, let $f:\Bbb R^2\to\Bbb R:\langle x,y\rangle\mapsto x+y$; then the open sets in $\langle\Bbb R^2,\tau\rangle$ are precisely the sets $f^{-1}[U]$ such that $U$ is an open set in the usual topology on $\Bbb R$. In fact, 
$$T_\epsilon(x)=f^{-1}[(f(x)-\epsilon,f(x)+\epsilon)]\;.$$
A: I agree that it is a basis for a topology on $\mathbb{R}^2$. It is not the standard topology. I think this should be homeomorphic to the quotient of $\mathbb{R}^2$ (with the standard topology) by the relation $x \sim y$ iff $x_1 + x_2 = y_1 + y_2$.
A: The basis sets are $B_{\alpha,\epsilon}=\{x | |x_1+x_2 -\alpha| < \epsilon \}$, with $\alpha \in \mathbb{R}$ and $\epsilon>0$.
Let $\phi(x) = (x_1+x_2,x_2-x_1)$. We see that $\phi$ is bijective.
Note that $\phi(B_{\alpha,\epsilon}) = \{ x | |x_1-\alpha| < \epsilon \} = (\alpha-\epsilon, \alpha+\epsilon) \times \mathbb{R}$.
It is easy to see that these sets form a basis for a topology, and since $\phi$ is a bijection, the sets $B_{\alpha,\epsilon}$ form a basis for a topology.
