Prove that the usual topology is the coarsest topology on $\mathbb{R}^2$ such that the addition $\sigma(x,y)=x+y$ and subtraction $\delta(x,y) = x-y$ maps are continuous maps to $\mathbb{R}$ (with the usual topology).
I know that if I take the preimage of an open interval in $\mathbb{R}$, I get a figure that looks sort of like an infinite parallelogram, but I don't see how I can proceed to show that the usual topology is the coarsest.
Best Answer
The preimage wrt $+$ of an interval is a diagonal strip with negative slope, while the preimage wrt $-$ of an interval is a diagonal strip with positive slope. If a topology on $\mathbb{R}^2$ has both $+$ and $-$ as continuous functions, all such diagonal strips must be open (this is necessary and sufficient), and in particular their intersections (proper parallelograms) must also be open.
We want the coarsest topology that satisfies these requirements, meaning we don't want any extra open sets. This tells us that the coarsest topology $\tau_c$ which satisfies our requirements is the one with a basis of open diagonal strips.
Now we have to show that this new topology is equal to the usual one, $\tau_u$. $\tau_c \subseteq \tau_u$ is straightforward, as each basis element is just an open strip (which is easily seen to be open in $\tau_u$). $\tau_u \subseteq \tau_c$ isn't too much harder. For any open ball and any point, we can find a smaller parallelogram (the intersection of open strips) which contains the point and is contained in the open ball. The open ball is then the union of all such paralellograms, meaning the usual topology is "generated" by $\tau_c$. Thus, $\tau_c = \tau_u$