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A subspace $Y$ of a topological space $X$ is disconnected exactly when it can be written as a union $Y=A\cup B$ with $A$ and $B$ disjoint nonempty open subsets of $Y$ (in the subspace topology). In terms of the topology of $X$, we have $A=U\cap Y$ and $B=V\cap Y$ for open sets $U$ and $V$ in $X$. This translates to:

$(1)$ $Y$ is disconnected iff $Y\subset U\cup V$ for some open sets $U$ and $V$ in $X$, both meeting $Y$, and with $U\cap Y$ and $V\cap Y$ disjoint.

As indicated in Willard, Exercise 26D, the following is not true in general:

$(2)$ $Y$ is disconnected iff $Y\subset U\cup V$ for some disjoint open sets $U$ and $V$ in $X$, both meeting $Y$.

We cannot take $U$ and $V$ disjoint in condition $(1)$ in general. See the answer to this question for a $T_0$ example. A simple $T_1$ example is the line with two origins $X=\mathbb{R}\cup\{0^*\}$. The subspace $Y=\{0,0^*\}$ is disconnected, with the only possible disconnection being $\{0\}\cup\{0^*\}$, but any two respective nbhds of $0$ and $0^*$ in $X$ are not disjoint.

However, if $X$ is a metric space, we can require $U$ and $V$ to be disjoint, as shown here.

Question:

Is there an example of a Hausdorff space $X$ with a disconnected subspace $Y$, where the disconnection cannot be witnessed by the traces on $Y$ of disjoint open sets in the ambient space?

As mentioned above, any counterexample cannot be a metrizable space. Also $Y$ cannot be finite, as any two points can be separated by disjoint open sets in a $T_2$ space. I just can't come up with any example.

PatrickR
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  • Suppose $Y=C\cup D$ where $C,D$ are disjoint non-empty closed connected subspaces of $X$ that are not completely separated in $X,$ that is, if $U,V$ are open in $X$ with $C\subset U$ and $D\subset V$ then $U\cap V$ is not empty. – DanielWainfleet Aug 09 '21 at 09:06

1 Answers1

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With respect to the usual topology on $\Bbb R^2,$ for $n\in \Bbb N$ let $C_n$ be the closed semi-circular arc in $\Bbb R\times [0,\infty)$ joining $(1/n,0)$ to $(1/(n+1),0).$

Let $X=(\,[0,\infty)\times \{0\}\,)\cup (\cup_{n\in\Bbb N}C_n\,).$

Let $T$ be the usual topology on $X$ as a subspace of $\Bbb R^2$. Let the topology $T_X$ be the weakest topology $T'$ on $X$ such that $T'\supset T$ and such that $\cup_{n\in\Bbb N}C_n$ is $T'$-closed. Let $C=\cup_{n\in\Bbb N}C_n$ and let $D=\{(0,0)\}$ and let $Y=C\cup D.$

Details are for the reader. Hints: (1). $C$ and $D$ are connected closed subspaces of $X$. (2). If $U,V\in T_X$ with $C\subset U$ and $D\subset V$ then $(U\cap V)\cap (X\setminus Y)\ne \emptyset.$

DanielWainfleet
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  • I thought about the topology $T^$ on $\Bbb R$ generated by $T_{\Bbb R}\cup {\Bbb R\setminus {1/n:n\in \Bbb N},}$.... which is $T_2$ but not $T_3$, because ${0}$ and the $T^$-closed set ${1/n:n\in \Bbb N}$ cannot be completely $T^*$-separated. I modified this so that $C$ (in my answer) is a connected subspace) – DanielWainfleet Aug 09 '21 at 09:46
  • In my above comment to my answer, $T_{\Bbb R}$ is the usual topology on $\Bbb R$. – DanielWainfleet Aug 09 '21 at 10:00
  • Very nice example. – PatrickR Aug 09 '21 at 20:08
  • Just repeating what you explained, examples can generally be found in Hausdorff spaces $X$ that are not regular. We can find a point $a$ and a closed subset $B$ not containing $a$ such that they cannot be separated by disjoint open sets in $X$. As long as $B$ is connected, the subspace $Y=B\cup{a}$ will be a counterexample. Thanks again. – PatrickR Aug 09 '21 at 20:24