A subset $Y$ of a topological space $X$ is disconnected if the subspace topology on $Y$ is disconnected. I'll define $Y$ to be 'disconnected in $X$' if there exists open $U,V$ (in $X$) such that $$ U \cup V \supseteq Y \qquad \mathrm{and} \qquad U \cap V = \varnothing \qquad \mathrm{and} \qquad U \cap Y \neq \varnothing \neq V \cap Y $$ In general, a disconnected subset $Y$ is not necessarily disconnected in $X$. As this question shows, we may be able to engineer a topology on $X$ that ensures any two open sets in $X$ that respectively contain the two disconnected components of $Y$ in the subspace topology always intersect, somewhere outside $Y$. To repeat the example, consider: $$ Y = \{1,2\} \qquad X = \{1,2,3\} \qquad T_X = \{\varnothing, \{3\} ,\{1,3\},\{2,3\}, X\} $$ Then $T_Y$ is discrete and disconnected, but there are no two open sets in $X$ that satisfy the requirements above.

My question, then, is: are there certain conditions on $X$ that ensure any disconnected subset of $X$ is 'disconnected in $X$'? For instance, a disconnected subset $Y$ will be 'connected in $X$' if the two disconnected components (taken as subsets of open sets in $X$) always intersect somewhere outside the subset, but it seems to me that this will force the space to be non-Hausdorff. However, I couldn't prove this was true, and my intuition for topology is not very good, so I wanted to ask a) whether disconnected subsets of Hausdorff spaces are always 'disconnected in $X$' and b) if not, whether there was some other class of topological space for which this was true.

Thank you.

  • $\begingroup$ This would be an example of what is called a "separation axiom" on $X$. A space being normal is the same as your property being true for any $Y$ that is the union of two disjoint, closed subsets of $X$, while Hausdorff is for $Y$ consisting of two points. Your property in general is strictly stronger than both normal and Hausdorff, however, since any metric space is both normal and Hausdorff, while for instance $X = \Bbb R$ with $Y = X \setminus \{0\}$ (and the standard topology) does not fulfill your condition. $\endgroup$ – Arthur Aug 2 '16 at 22:07
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    $\begingroup$ @Arthur: If $Y$ is open in $X$, then $U\cap Y$ and $V\cap Y$ are open, and so replacing $U$ and $V$ by $U\cap Y$ and $V\cap Y$, we can ensure that they have empty intersection in $X$. So I'm not sure about your last sentence. Or have I misunderstood? $\endgroup$ – tracing Aug 2 '16 at 22:49
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    $\begingroup$ @tracing That was not a correct example, you're right. Still, since this condition does not limit what $U$ and $V$ can be, it's hard to imagine that it's equivalent to normal, so I am certain that there is some counterexample out there. Apparently, I'm just too tired to think up pathological examples right now. $\endgroup$ – Arthur Aug 2 '16 at 22:53
  • $\begingroup$ From your argument we can certainly conclude that my condition is at least as strong as the two separation axioms mentioned, right? And it's definitely believable to me that if Hausdorff is equivalent to my condition being true just for subsets consisting of two points, then my condition holding for any subset is probably a stronger requirement. Is this reasonable do you think? $\endgroup$ – gj255 Aug 2 '16 at 23:09
  • $\begingroup$ Your property does imply that $X$ is normal, but it does not imply that $X$ is Hausdorff: consider the indiscrete topology. It does, however, imply that if $X$ is $T_1$, so that singletons are closed, then $X$ is Hausdorff. $\endgroup$ – Brian M. Scott Aug 3 '16 at 0:53

Your property is equivalent to complete normality.

Recall that subsets $A$ and $B$ of $X$ are said to be separated if $A\cap\operatorname{cl}B=\varnothing=B\cap\operatorname{cl}A$, and $X$ is completely normal if every pair of separated sets in $X$ have disjoint open nbhds. $X$ is $T_5$ if $X$ is completely normal and $T_1$.

Suppose that $X$ is completely normal, and let $Y$ be a subspace of $X$ that is not connected. Then there are open sets $U_0$ and $U_1$ in $X$ such that if $G_0=U_0\cap Y$ and $G_1=U_1\cap Y$, then $\{G_0,G_1\}$ is a partition of $Y$ into (non-empty) relatively open sets. $U_0$ is an open nbhd of $G_0$ disjoint from $G_1$, so $G_0\cap\operatorname{cl}_XG_1=\varnothing$. Similarly, $G_1\cap\operatorname{cl}_XG_0=\varnothing$, so $G_0$ and $H_0$ are separated subsets of $X$. $X$ is completely normal, so there are disjoint open sets $V_0$ and $V_1$ such that $G_0\subseteq V_0$ and $G_1\subseteq V_1$. This shows that every completely normal space has your property.

Conversely, suppose that $X$ has your property, and let $A$ and $B$ be separated sets in $X$. Then $A$ and $B$ are disjoint, non-empty, relatively closed subsets of $A\cup B$, which is therefore not connected. Since $X$ has your property, $A$ and $B$ have disjoint open nbhds in $X$, and $X$ is therefore completely normal.

By the way, complete normality is equivalent to hereditary normality: all subspaces are normal.

Note that every space with the indiscrete topology is vacuously completely normal and therefore has your property, so your property does not imply that the space is Hausdorff. If $X$ is $T_1$ and has your property, however, then singletons are closed, and $X$ must be Hausdorff.

  • $\begingroup$ Since metric spaces are normal, and being a metric space is hereditary, this means that metric spaces are $T_5$, right? If so, maybe this could be mentioned in your answer, since metric spaces come up quite a bit in practice? (And apologies if you don't like the suggestion, or if I'm getting muddled.) $\endgroup$ – tracing Aug 3 '16 at 22:33
  • $\begingroup$ @tracing: Yes, metric spaces are $T_5$; so are linearly ordered spaces with the order topology, though that, unlike the complete normality of metric spaces, isn't obvious. $\endgroup$ – Brian M. Scott Aug 3 '16 at 22:39

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