I guess there's something wrong with my thoughts about connectedness seen in a subspace of a topological space and I need your help. Let me explain:

These are the definitions I have:

A topological space $X$ is said to be disconnected if there are disjoint nonempty open sets $U, V\subset X$ such that $X=U\cup V$. A subspace $Y$ of $X$ is disconnected if it is disconnected considered as a space endowed with the relative topology to $Y$.

But I have just come across this exercise in Willard's General Topology that says:

Among the criteria for a subspace $E$ of $X$ to be connected, the following was absent: $E\subset X$ is disconnected iff there are disjoint open sets $H$ and $K$ in $X$, each meeting $E$, such that $E\subset H\cup K$. Find a counterexample.

This really confuses me and I haven't found any counterexample yet. If we had such sets $H$ and $K$, we could then consider $E_H=E\cap H$ and $E_K=E\cap K$, and have $E=E_H\cup E_K$. Therefore, $E$ is disconnected because we can express it as the union of a pair of disjoint nonempty open sets relative to the subspace topology.

I can't see what is wrong about it, but there must be a mistake somewhere.


Note that "iff" means implications both ways. Disproving an iff statement only requires showing that one of the implications does not always hold. You have indicated why that condition does imply disconnectedness. So the question is, can we find a disconnected subspace of a topological space that doesn't satisfy that condition?

I suggest thinking about the cofinite topology on an infinite set.

  • $\begingroup$ Ok. In that case, I know finite sets are disconnected. So if there were open sets $U,V$, their complement would be finite, hence one of them wouldn't be actually open, right? $\endgroup$ – user55334 Jan 6 '13 at 9:56
  • $\begingroup$ @user55334: In a cofinite topology, yes, finite subspaces with more than one element are disconnected (and even discrete). And because nonempty open sets in the larger space always intersect, the condition won't hold. $\endgroup$ – Jonas Meyer Jan 6 '13 at 10:02
  • $\begingroup$ Thank you. Everything's clear now. $\endgroup$ – user55334 Jan 6 '13 at 10:04
  • $\begingroup$ @user55334: You're welcome. I'm glad. No. $\endgroup$ – Jonas Meyer Jan 6 '13 at 10:05
  • $\begingroup$ Alright, no problem. $\endgroup$ – user55334 Jan 6 '13 at 10:06

Another example: Consider the set $\{a,b,c\}$ with topology $\{a,b,c\},\{\},\{a,b\},\{b,c\},\{b\}$. Then the subspace $\{a,c\}$ is disjoint because $\{a\}$ and $\{c\}$ are open sets in this subspace. However, in the original space, every open set containing $a$ and every open set containing $c$ necessarily contain $b$, so they cannot be disjoint.


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