# Prove or disprove : $(\mathbb R,\tau)$ is connected and Hausdorff?

Consider the topology $\tau = \{ G \subseteq \mathbb R: G^c\text{ is compact in }(\mathbb R, \tau_u) \} \cup \{\mathbb R, \varnothing\} \text{ on } \mathbb R$ , where $\tau_u$ is the usual topology on $\mathbb R$ . Then $(\mathbb R , \tau)$ is

1. a connected Hausdorff space

2. connected but not Hausdorff

3. Hausdorff but not connected

4. neither Hausdorff nor connected

My attempt is :

If $G_1 , G_2 \in \tau$ such that $G_1 \cap G_2 = \varnothing$ , then

$G_1^c \cup G_2^c = \mathbb R$ so $\mathbb R$ is compact because union of two compact set is compact , which is a contradiction . so $(\mathbb R, \tau)$ is not Hausdorff.

Please tell me about the connectedness .

Thank you

## 3 Answers

Hint: Connectedness is equivalent to the assertion that the only subsets which are both open and closed are $\emptyset$ and $\mathbb{R}$.

Does $\tau$ contain two nonempty disjoint open sets whose union is all of $\mathbb{R}$? The argument you've already given is on the right track to attack this.

Hint If $A$ is unbounded, then the closure of $A$ in $(\Bbb{R},\tau)$ is $\Bbb{R}$. Note that the space $X$ is connected iff there is no $A$, $B$ satisfy that $A\cup B=X$, $\overline{A}\cap B = A\cap\overline{B}=\varnothing$.

• @tetory : Thanku for your prompt reply, but how to show your initial point : Let A is unbounded , x $\in \mathbb R$ \ A then $\exists$ G in $\tau$ such that $A\cap G = \phi$ $\Rightarrow A^c \cup G^c = \mathbb R$ – Struggler Mar 3 '14 at 5:39
• @user112064 $G$ is open in $(\Bbb{R},\tau)$ iff $G^C$ is bounded and closed in $(\Bbb{R},\tau_u)$. Especially $G^C$ is bounded. So, if $x\in A^C$ and $G\in\tau$ is a neighborhood of $x$ then $A\cap G$ is not empty, since $A$ is unbounded and $G^C$ is bounded. – Hanul Jeon Mar 3 '14 at 5:45
• More precisely, If $G^C\subset [-M,M]$ for some $M>0$ and $A$ is unbounded then $A\cap [-M,M]^C$ is not empty. – Hanul Jeon Mar 3 '14 at 5:46