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 $R$ with topology $T$ on $R$ consisting of all sets of the form $(x,\infty)$, where $x\in R$, is not a $T_3$ space.
I have $(0,\infty)$ is open in this space. So, $A=R-(0,\infty)=(-\infty,0]$ is closed in this space and $1\notin (-\infty,0]$. Suppose there exist $U,V$ open in $R$ such that $A\subset U$ and $1 \in V$. I claimed that $U \cap V=\emptyset$. I don't see how an open set in $R$ can contain $A$ but does not contain $1$ because any open set in $R$ must be of the form $(x,\infty)$. 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 $R$ with topology $T$ on $R$ consisting of all sets of the form $(x,\infty)$, where $x\in R$, is not a $T_3$ space.
I have $(0,\infty)$ is open in this space. So, $A=R-(0,\infty)=(-\infty,0]$ is closed in this space and $1\notin (-\infty,0]$. Suppose there exist $U,V$ open in $R$ such that $A\subset U$ and $1 \in V$. I claimed that $U \cap V=\emptyset$. I don't see how an open set in $R$ can contain $A$ but does not contain $1$ because any open set in $R$ must be of the form $(x,\infty)$. 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 $U$ of $0$ contains $1$ since there must exist a set of the form $(a,\infty)\subseteq U$ with $a<0$ but then $1\in (a,\infty)\subseteq U$