1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $A\cup B=X$, $\overline{A}\cap B = A\cap\overline{B}=\varnothing$.

$\endgroup$
  • $\begingroup$ @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$ $\endgroup$ – Struggler Mar 3 '14 at 5:39
  • $\begingroup$ @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. $\endgroup$ – Hanul Jeon Mar 3 '14 at 5:45
  • 1
    $\begingroup$ 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. $\endgroup$ – Hanul Jeon Mar 3 '14 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.