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I want to show that $\tau=\{G_k=(x,y):k\in \mathbb R\}\cup \{\mathbb{R}^2, \emptyset\}$, where $\forall k\in \mathbb R, \;\; G_k=\{(x,y):x>y+k\}$ is a topology on $\mathbb R^2$.

My attemp: By definition $\mathbb R^2,\emptyset\in \tau$. Let $\{G_\alpha:\alpha\in \Lambda\subset \mathbb R\}$ be any family of subsets of $\mathbb R^2$ (To avoid trivial case let us assume that $\mathbb R^2$ is not there). If $\sup \Lambda=k\in \mathbb R$, then $\bigcup\limits_{\alpha \in \Lambda}G_\alpha=G_k$ and if $\sup \Lambda=\infty$, then $\bigcup\limits_{\alpha \in \Lambda}G_\alpha=\mathbb R^2$.

If we consider $G_k,G_l\in \tau$, then $G_k\cap G_l=G_i$, where $i=\min\{k,l\}$.

Please tell me if I am right or show the flaw in the proof. Thanks in advance.

user100943
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  • You should show that $G_{\alpha} \subset G_{\beta}$ under the right conditions – Daniel Donnelly Oct 18 '13 at 17:20
  • @user 100943: I do not think this is correct. If $(x,y)\in \bigcup\limits_{k\in \Lambda}G_k$, then how is it possible that $(x,y)\in G_l$ if $l=\sup \Lambda$? – Anupam Oct 19 '13 at 03:21

1 Answers1

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You are approximately right for the union.All you have to do is to prove what you say. Firstly,let's consider a bounded set Λ in $\Bbb R$.

Then there exists a strictly increasing sequence $({z_n})\subset Λ $:$z_n-> supΛ$ .<=> for every ε>0 there exists a $k(ε)\in\Bbb N$:|$z_n$ - supΛ|<ε for every $n>=k$. Then for every $ε>0$ there exists a $k(ε)\in \Bbb N$:$G_{z_n}$=$G_{supΛ-ε}$ for every $n>=k$. (1)

Also we have that $G_a < G_b \iff a < b$. So $G_{z_n}$ < $G_{z_{n+1}}$.(2)

From (1),(2) we have that $\bigcup\limits_{\alpha \in \Lambda}G_\alpha=G_(supΛ-ε)$ for an $ε \in \Bbb R$ every time. Thus $\bigcup\limits_{\alpha \in \Lambda}G_\alpha\in T$.

Same we do for the trivial case that Λ is not bounded and the union is $\Bbb R^2$.

Haha
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