0
$\begingroup$

Let $X$ and $X'$ denote the same set under different topologies $\tau $ and $\tau'$ where $\tau' \supset \tau$. I have to decide what one of them would be like if the other one is like Regular or Normal spaces.

So , first I consider $X=\mathbb R=X'$ and $\tau'$=euclidean topology and $\tau$= co-finite topology

Then easily $X'$ is both normal and regular but $X$ is neither.

Going the other way round is a little problem. For the given condition tells that any open set in $\tau$ is open in $\tau'$ but the converse may not hold.

Now if I assume that $X$ is Regular. If $p'$ is a point in $X'$ disjoint from the closed set $C'.$ If I want to prove that $X'$ is Regular then I have to find open sets $U'$ and $V'$ in $X'$ such that $$U'\cap V'=\varnothing$$ and $$p'\in U' \\C'\subset V'$$ . This would be done if only I could tell that $C=C'$[in $X$] were closed in $X$.

How can I , if at all , prove that $?$ Or if that is not the fact then I need some counter-example to disprove that statement .

Please help me with the $Regular$ , I will try to do the $Normal$ on my own .

Thanks.

$\endgroup$
2
  • $\begingroup$ Do your definitions of "regular" and "normal" include that points are closed? (There are counterexamples either way, but there are much simpler counterexamples if you do not require points to be closed.) $\endgroup$ Nov 24 '15 at 9:03
  • $\begingroup$ @EricWofsey : Yes , sir . Singleton points are closed. $\endgroup$
    – user118494
    Nov 24 '15 at 10:02
2
$\begingroup$

The following is a well-known and handy counterexample. Let $\tau$ denote the Euclidean topology on $\mathbb{R}$ and let $A=\{1/n:n\in\mathbb{Z}_+\}\subset\mathbb{R}$. Define $$\tau'=\{U\subseteq\mathbb{R}:U\cup B\in\tau\text{ for some }B\subseteq A\}.$$

I will let you verify the following facts about $\tau'$:

  1. $\tau'$ is a topology on $\mathbb{R}$.

  2. $\tau\subset\tau'$.

  3. $A$ is a closed set with respect to $\tau'$.

  4. There do not exist $U,V\in\tau'$ such that $A\subseteq U$, $0\in V$, and $U\cap V=\emptyset$.

In particular, (3) and (4) together imply that $\tau'$ is not regular.

$\endgroup$
1
  • $\begingroup$ So , I can use this space as a counter example for the $Normal$ case as well since $Normal$ implies $Regular$ . Also ${\mathbb R_l}^2$ with lower limit topology $\ [$ Sorgenfrey Plane $]\ $ is not normal but $\mathbb R^2$ under usual topology is . So , that can also be a counter example for the Normal case as lower-limit-topology is finer than the euclidean-topology.$[\ $ Fineness passing onto finite product $]\ $ Correct $?$ $\endgroup$
    – user118494
    Nov 24 '15 at 11:49

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.