Can an open set be covered by proper open subsets? Let $(X, \tau)$ be a topological space, let $U\in\tau$. An open cover of $U$ is a set $\{U_i \ |\ i\in I\}$ (of open sets $U_i$) whose union $\bigcup U_i$ contains $U$.
If $U\subsetneq X$, then $U$ admits an open covering by open sets $\{U_i \ |\ i\in I\}$, where an arbitrary open set $U_i$ may not be a subset of  $U$. (I think.)
Does $U$ admit an open covering by open sets $\{U_i \ |\ i\in I\}$ where every open set $U_i$ is a proper subset of $U$? Less formally: given an arbitrary open set $U$, can we find an open cover of $U$ made up only of open sets "inside" of $U$?
If the answer is no, then what are the open sets $U$ that admit such "internal" coverings?
Is the answer any different if we replace $U$ by an arbitrary closed set?

A space is called $T_1$ if, for any two points in the space, each has an open neighborhood that misses the other. This is one of many separation axioms that measure how strongly we can separate points in the space. For some mathematicians, topological spaces aren't even worth considering until they are at least $T_2$ (Hausdorff), which is even stronger. (This is to say that all but the most pathological topologies are at least $T_1$.)
We can construct a cover of the type you describe if the space is at least $T_1$ and $U$ contains more than one point. For each $x \in U$, construct an open neighborhood $U_x$ as follows:


*

*Choose any $y \in U$ that is distinct from $x$.

*Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.

*Set $U_x = G_x \cap U$, so that $U_x$ is a proper open subset of $U$ containing $x$.


Finally, we can take $\{U_x \mid x \in U\}$ as an open cover of $U$ by proper open subsets.
