# Connected subspaces

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.

• 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? – user55334 Jan 6 '13 at 9:56
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.