Let $U(a,t) = \{a\} \cup [t, \infty)$ where $a,t \in \Bbb R.$ Show that the generated space is not Hausdorff.

Question

Let $U(a,t) = \{a\} \cup [t, \infty)$ where $a,t \in \Bbb R.$ The collection of these sets forms a basis for some topology $\tau$ on $\Bbb R$. Show that the generated space is not Hausdorff.

Let $x,y \in \Bbb R$. We need to show that $O_x \cap O_y = \emptyset$. I am confused about what are these sets $O_x$ and $O_y$? Are they just unions and intersections of the basis elements $U(a,t)$ or how should I express them? I don't think I have $O_x=U(x,t)$ and $O_y=U(y,s)$ for $t, s \in \Bbb R$?

Even if I had the intersection $$(\{x\} \cup [t, \infty)) \cap (\{y\} \cup [s, \infty)) $$ is neve non-empty as it's at least one of the intervals?

Answer 1

Let $N_0$ and $N_1$ be neighborhoods of $0$ and $1$ respectively. Then $N_0\supset U(a,t)$ and $N_1\supset U(b,s)$ for some numbers $a$, $b$, $s$, and $t$ such that $0\in U(a,t)$ and that $b\in U(b,s)$. But then $N_0\cap N_1\ne\emptyset$, since$$\max\{s,t\}\in U(a,t)\cap U(b,s)\subset N_0\cap N_1.$$

Answer 2

Let $x,y \in \Bbb R$. We need to show that $O_x \cap O_y = \emptyset$.

No, that is not what you need to show.

Remember to always follow the definitions. You need to show that the space is not Hausdorff.

Remember, by definition, a space $(X, \tau)$ is Hausdorff if, for every pair of points $x,y\in X$, there exist two open sets $O_x$ and $O_y$ such that $x\in O_x, y\in O_y$ and $O_x\cap O_y=\emptyset$.

In your case, you must prove the space is not Hausdorff, and you can do that by proving the above statement is false, or in other words, you must find some pair of points $x,y\in\mathbb R$ so that every open neighborhood of $x$ must intersect every open neighborhood of point $y$.