I was wondering how to find a topology on $\mathbb{R}$ and a disconnected subset $A$ such that every pair of sets, $U$ and $V$, that is a separation of $A$ in $\mathbb{R}$ satisfies $U \cap V \cap (\mathbb{R}- A ) \ne \varnothing$?
I was thinking the lower limit topology on $\mathbb{R}$ would be such a topology.
For example, if $\mathbb{R}= (-\infty, \infty), A = (-\infty,0), U = (-\infty, 0)\cup[1,\infty), V = [0,\infty)$.
Then
- $A\subset U\cup V$
- $A\cap U\ne\varnothing$
- $A\cap V\ne\varnothing$
- $U\cap V\cap A=\varnothing$
So $U$ and $V$ are indeed a seperation of $A$ and $A$ is disconnected in $\mathbb{R}$.
And also we have $U\cap V\cap(\mathbb{R}-A)\ne\varnothing$, which is what we wanted.
But I am unsure if that works in all cases. We need to show this holds for all $U$ and $V$ pairs that are a separation of $A$.
Any help would be much appreciated! Thanks.
(Getting another question off the unanswered list.)
The lower-limit topology does not provide an example. Let $A$ be any subset of $\Bbb R$ with at least two points, and fix $a,b\in A$ with $a<b$. Let $U=(\leftarrow,b)$ and $V=[b,\to)$; then $U$ and $V$ are disjoint open sets in $\Bbb R$ with the lower-limit topology, $a\in U\cap A$, and $b\in V\cap A$, so $\{U,V\}$ is a separation of $A$ such that $U\cap V=\varnothing$ (and hence of course $U\cap V\cap(\Bbb R\setminus A)=\varnothing$).
user647486 gave a correct example in the comments, but in terms of separation properties it isn’t very nice, since it fails badly even to be $T_0$. It actually isn’t hard to find a $T_1$ example: the cofinite topology on $\Bbb R$ works nicely. Let $A=\{a,b\}\subseteq\Bbb R$, where $a\ne b$, let $U\subseteq\Bbb R\setminus\{b\}$ be an open nbhd of $a$ in the cofinite topology, and let $V\subseteq\Bbb R\setminus\{a\}$ be an open nbhd of $b$ in the cofinite topology; then $\{U,V\}$ is a separation of $A$, and every separation of $A$ has this form. By definition there are finite sets $F,G\subseteq\Bbb R$ such that $U=\Bbb R\setminus F$ and $V=\Bbb R\setminus G$; then $U\cap V=(\Bbb R\setminus F)\cap(\Bbb R\setminus G)=\Bbb R\setminus(F\cup G)\ne\varnothing$. Moreover, $a\in G$ and $b\in F$, so $U\cap V\cap(\Bbb R\setminus A)=U\cap V\ne\varnothing$.
In fact with some work one can construct a compact Hausdorff topology on $\Bbb R$ with the desired property. Note first that it suffices to find a compact Hausdorff space $\langle X,\tau_X\rangle$ with a non-connected subset $A$ such that $|X|=|\Bbb R|$, and if $\{U,V\}$ is any separation of $A$ in $X$, then $U\cap V\cap(X\setminus A)\ne\varnothing$: if $\varphi:X\to\Bbb R$ is any bijection, then $\tau=\{\varphi[U]:U\in\tau_X\}$ is a compact Hausdorff topology on $\Bbb R$, $\varphi[A]\subseteq\Bbb R$ is not $\tau$-connected, and if $\{U,V\}\subseteq\tau$ is any separation of $\varphi[A]$, then $U\cap V\cap(\Bbb R\setminus\varphi[A])\ne\varnothing$.
Let $L$ be the closed long ray: $L=\omega_1\times[0,1)$ with the order topology generated by the lexicographic order $\preceq$ on $\omega_1\times[0,1)$. Let $p$ be a point not in $L$, let $L^*=L\cup\{p\}$, and extend $\preceq$ to $L^*$ by letting $x\prec p$ for each $x\in L$; $L^*$ with the resulting order topology is a compact Hausdorff space, and $|L^*|=|\Bbb R|$. ($L^*$ is in fact the one-point compactification of $L$.)
Let $X=L^*\times[0,1]$ with the product topology; $X$ is a connected variant of the Tikhonov plank. Let $C=L\times\{1\}$, $D=\{p\}\times[0,1)$, and $A=C\cup D$. Then $X$ is compact and Hausdorff, $|X|=|\Bbb R|$, and $\{L\times[0,1],L^*\times[0,1)\}$ is a separation of $A$ into parts $C$ and $D$, so $A$ is not connected. Suppose that $\{U,V\}$ is a separation of $A$; without loss of generality we may assume that $U\cap C\ne\varnothing$, and since $C$ is connected, we must have $C\subseteq U$. $D$ is also connected, so we must have $D\cap U=\varnothing$ and hence $D\subseteq V$.
Let $\mathscr{B}$ be a countable base for $[0,1)$; for each $r\in[0,1)$ there are a $B_r\in\mathscr{B}$ and an $x_r\in L$ such that
$$\langle p,r\rangle\in (x_r,p]_\preceq\times B_r\subseteq V\,.$$
Let $\mathscr{B}_0=\{B_r:r\in[0,1)\}$, and for each $B\in\mathscr{B}_0$ fix $r(B)\in[0,1)$ such that $B_{r(B)}=B$; then
$$D\subseteq\bigcup_{B\in\mathscr{B}_0}\big((x_{r(B)},p]\times B\big)\subseteq V\,.$$
Now $\{x_{r(B)}:B\in\mathscr{B}_0\}$ is countable, so there is a $y\in L$ such that $x_{r(B)}\preceq y$ for each $B\in\mathscr{B}_0$, and it follows immediately that
$$D\subseteq(y,p]\times[0,1)=\bigcup_{B\in\mathscr{B}_0}\big((y,p]\times B\big)\subseteq V\,.\tag{1}$$
Fix $z\in(y,p)$. $U$ is an open nbhd of $C$, and $\langle z,1\rangle\in C$, so there is some $s\in[0,1)$ such that $\langle z,s\rangle\in U$. By $(1)$ we have $\langle z,s\rangle\in V$, and clearly $\langle z,s\rangle\notin A$, so $U\cap V\cap(X\setminus A)\ne\varnothing$, and we’re done.