I will try to prove $\Bbb R$ is regular
Attempt (edited)
Let $\Bbb R$ be defined on usual topology
Let A be closed and A$\subset\Bbb R$ and x$\in \Bbb R$ and x$\notin A$
So if [a,b] is closed then it’s complement U=$\Bbb R\setminus [a,b]$ is open.
By Def 3.2.1 if x$\in\Bbb R $ then x$\in U\subset \Bbb R $ s.t (a,b)$\cap A=\emptyset$
And U=(-$\infty$,a)$\cup(b,\infty$) which is the union of open sets,so open
So U $\supseteq A$ and A and
(-$\infty$,a) and (b,$\infty$)are disjoint containing x and A respectively ?
Any help would be appreciated.
Am l missing anything?
Do l have to find to subsets within (-$\infty$,a) and (b,$\infty$)
Best Answer
As I said in my comment, you are supposed to assume that $\Bbb R$ has its usual topology, not the finite-closed topology (also known as the cofinite topology), and most of what you’ve written suggests that you were in fact thinking of the usual topology. For instance, an interval $[a,b]$ is not finite unless $a=b$, so in general it is not closed in the finite-closed topology, but it is closed in the usual topology. However, it is very far from being the only kind of closed set: $\Bbb Z$ is a closed subset of $\Bbb R$ that is not of the form $[a,b]$, as is $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, to name just two.
Now let’s take a look at your argument. You start with a set $A\subseteq\Bbb R$ and a point $x\in\Bbb R$ such that $x\notin A$; since you’re trying to prove that $\Bbb R$ is regular, that’s a very reasonable place to start, except that you also want to assume that $A$ is closed. After that, though, things go a bit haywire. You let $p$ be a point of $A$, but why? You never mention $p$ again, so it serves absolutely no purpose.
Then you say that we can find ‘an open nbhd of $[a,b]$ of $x$ such that $[a,b]\cap A=\varnothing$’. I suspect that the first of wasn’t supposed to be there, and it was supposed to read ‘an open nbhd $[a,b]$ of $x$’, but $[a,b]$ isn’t open. Its complement $U=(\leftarrow,a)\cup(b,\to)$ is open in the usual topology, not the ‘closed topology’, what ever that is.
As you’ve set things up it is indeed true that $U\supseteq A$, and it is also true that $(\leftarrow,a)$ and $(b,\to)$ are disjoint, but $A$ could be a subset of either of those rays: there is no reason to think that $A\subseteq(b,\to)$. And since you apparently wanted $[a,b]$ to be a nbhd of $x$, $x$ must be in $[a,b]$, so clearly $(\leftarrow,a)$ cannot contain $x$.
Let’s go back to the beginning and start over. Let $A$ be a closed set in $\Bbb R$, and let $x\in\Bbb R\setminus A$. Then $\Bbb R\setminus A$ is an open set containing $x$, so there is an open interval $(a,b)$ such that $x\in(a,b)\subseteq\Bbb R\setminus A$. Since $x\in(a,b)$, we know that $a<x<b$, so there are $c,d\in\Bbb R$ such that $a<c<x<d<b$. Let $U=(c,d)$ and $V=(\leftarrow,a)\cup(b,\to)$; clearly $U$ and $V$ are open sets, $x\in U$, and $U\cap V=\varnothing$. I’ll leave it to you to complete the proof by showing that $A\subseteq V$.