Proving $\mathbb{T=\{R^2,\varnothing\}}\cup\{G_k,k\in\mathbb{Q}\}$ is not a topology.

Question

Consider the class $\mathbb{T=\{R^2,\varnothing\}}\cup\{G_k,k\in\mathbb{R}\}$ of substes in the $\mathbb{R^2}$ plane. Where $$G_k=\{(x,y)|x>y+ k\}$$

I have already proven $\mathbb{T}$ is a topology, however, if $\mathbb{T=\{R^2,\varnothing\}}\cup\{G_k,k\in\mathbb{Q}\}$ it is not a topology anymore, why? I can't prove it.

Answer 1

Take a decreasing sequence $\mathbb{Q}\ni a_n\searrow \pi$ (slight modification of yours), and consider the family of sets $G_{a_n}$.

By definition $G_{a_n}\in \mathbb{T}$.If $\mathbb{T}$ were a topology, then the union of the $G_{a_n}$ would again be in the topology.

But $\bigcup G_{a_n}=G_{\pi}$ (is this clear to you?) and $G_\pi$ is not in $\mathbb{T}$ as $\pi\not\in\mathbb{Q}$.

Thus we have a contradiction and hence $\mathbb{T}$ is not a topology.