1. Regular space

I am working on this problem, but I am stuck in the middle. Could someone please give me a hand?
I am trying to prove that the set of real numbers$\displaystyle R$ with topology $\displaystyle T$ on $\displaystyle R$ consisting of all sets of the form $\displaystyle (x,\infty)$, where $\displaystyle x\in R$, is not a $\displaystyle T_3$ space.
I have $\displaystyle (0,\infty)$ is open in this space. So, $\displaystyle A=R-(0,\infty)=(-\infty,0]$ is closed in this space and $\displaystyle 1\notin (-\infty,0]$. Suppose there exist $\displaystyle U,V$ open in $\displaystyle R$ such that $\displaystyle A\subset U$ and $\displaystyle 1 \in V$. I claimed that $\displaystyle U \cap V=\emptyset$. I don't see how an open set in $\displaystyle R$ can contain $\displaystyle A$ but does not contain $\displaystyle 1$ because any open set in $\displaystyle R$ must be of the form $\displaystyle (x,\infty)$. However, I don't have a rigorous argument here.

2. Originally Posted by jackie
I am working on this problem, but I am stuck in the middle. Could someone please give me a hand?
I am trying to prove that the set of real numbers$\displaystyle R$ with topology $\displaystyle T$ on $\displaystyle R$ consisting of all sets of the form $\displaystyle (x,\infty)$, where $\displaystyle x\in R$, is not a $\displaystyle T_3$ space.
I have $\displaystyle (0,\infty)$ is open in this space. So, $\displaystyle A=R-(0,\infty)=(-\infty,0]$ is closed in this space and $\displaystyle 1\notin (-\infty,0]$. Suppose there exist $\displaystyle U,V$ open in $\displaystyle R$ such that $\displaystyle A\subset U$ and $\displaystyle 1 \in V$. I claimed that $\displaystyle U \cap V=\emptyset$. I don't see how an open set in $\displaystyle R$ can contain $\displaystyle A$ but does not contain $\displaystyle 1$ because any open set in $\displaystyle R$ must be of the form $\displaystyle (x,\infty)$. However, I don't have a rigorous argument here.
I think you are making this harder than it need be. This space isn't even Haudorff since any neighborhood $\displaystyle U$ of $\displaystyle 0$ contains $\displaystyle 1$ since there must exist a set of the form $\displaystyle (a,\infty)\subseteq U$ with $\displaystyle a<0$ but then $\displaystyle 1\in (a,\infty)\subseteq U$