Show that $\tau$ is a topology to $\mathbb{R}^2$. The problem is: Let $\tau=\{\mathbb{R}^2,\emptyset\}\cup\{G_k:k\in\mathbb{R}\}$ where $$G_k=\{(x,y)\in\mathbb{R}^2:x>y+k\}.$$Show that $\tau$ is a topology to $\mathbb{R}^2$. 
My attempt:

*

*Its obvious that $\mathbb{R}^2$ and $\emptyset$ are members of $\tau$.

*Let $\{A_i\}_{i\in I}\subseteq\tau$, if there is a $i_0\in I$ such that $A_{i_0}=\mathbb{R}^2$, then $\cup_{i\in I}A_i=\mathbb{R}^2\in\tau$ and we finish this part. In other case, let $k_i$ with $i\in I$ such that $A_i=G_{k_i}$ and $(x,y)\in\cup_{i\in I}A_i$, then $x>y+k_i$ for all $i\in I$, we can take $k=\max\{k_i:i\in I\}$ and all the inequalities are conserved. Then $\cup_{i\in I}A_i=G_k$ and for that, $\cup_{i\in I}A_i\in\tau.$

*Let $\{A_i\}_{i\in I_n}\subset\tau$. If there is a $i'$ such that $A_{i'}=\emptyset$ we have that $\cap_{i=1}^nA_i=\emptyset\subset\tau$ and we finish this part. In other case, if $A_i\neq\emptyset$ for al $i$, let $(x,y)\in\cap_{i=1}^nA_i$, then $(x,y)\in A_i$ for all $i$. There exists $k_1,...,k_n$ such that $x>y+k_i$ for all $i$, let $k=\max{k_1,...,k_n}$, we have that $x>y+k$ and all the inequalities are conserved, then $\cap_{i=1}^nA_i=G_k$ and then $\cap_{i=1}^nA_i\in\tau$.

Is my attempt correct? 
Thanks!!!
 A: Not quite. Let's analyze the part about the unions a bit. Let's consider $G_1$ and $G_2$ for example. Now $G_1$ is the region below the line $y=x-1$ while $G_2$ is the region below the line $y=x-2$. Their union of course is $G_1$. In general, $G_k$ is the region below the line $y=x-k$. This leads us to the conclusion that the union of finitely many $G_{k_i}$'s is actually $G_{\min k_i}$. So what's the problem with the infinite setting. The answer is that you can't guarantee that an infinite, let alone potentially uncountable, subset of $\mathbb{R}$ has a minimum. A concrete counterexample is the the family $\mathcal{A}=\{G_{-k}, k \in \mathbb{N}\}$. I'll leave it to you to verify that $\bigcup \mathcal{A}=\mathbb{R}^2$. So how can we fix that problem? We can distinguish two cases:

*

*The set $\{k_i: i \in I\}$ is bounded below. Then consider $m = \inf\{k_i: i \in I\}$ and show that $\bigcup_{i\in I}A_i = G_m$.

*The set $\{k_i: i \in I\}$ is not bounded from below. Then show that $\bigcup_{i\in I}A_i=\mathbb{R}^2$.

The first case might be a bit technical but should be doable. Finally, a piece of general advice for the intersections is that you can prove it for only two sets (which should make the solution a bit shorter). Apart from that though, your reasoning at 3. is correct
