# 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?

• Have you thought about the integers? – Steve D May 15 '17 at 21:38
• For many topological spaces you can do this. For example, any open subset of $\mathbb R^n$ is a union of open balls, which can be made proper. However, @SteveD's hint shows that you can't always do this. – Cheerful Parsnip May 15 '17 at 22:20
• If $X$ is a $T_1$ space and if $U$ is open and has more than $1$ member then $\{U$ \ $\{p\}: p\in U\}$ is an open cover of $U$ by open proper subsets of $U.$ ....If $\tau=\{X,\phi\}$ then with U=X, no proper non-empty subset of $U$ is open.... Another example is Sierpinski space $S=\{0,1\}$ with $\tau=\{S,\phi, \{0\}\}$. – DanielWainfleet May 15 '17 at 23:57
• If U is open, of course. If U is open every point is an interior point so for every point x in U there is an open neighborhood of x completely in U. Call this N_x, then $\cup_{x\in U} Nx$ is an open cover. If U is not open then of course not, If U is not open there is a point that is not an interior point and any open set containing it must contain points not in U (else it'd be an interior point). So all open covers contain points not in U. – fleablood May 16 '17 at 1:25
• @fleablood This is the argument I was going to give, but (as others pointed out to me) it only goes through if the space is at least $T_1$. – Austin Mohr May 16 '17 at 1:31

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:

1. Choose any $y \in U$ that is distinct from $x$.
2. Appeal to the $T_1$ property to get an open set $G_x$ that contains $x$ but does not contain $y$.
3. 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.

• But why is it guaranteed that the "first" $y$ we chose is an element of some element of $\{ U_x \ | \ x\in U \}$ ? Moreover, why is every such $y$ eventually in some of the cover elements? Because, by construction, $y$ is not in $U_x$ – étale-cohomology May 17 '17 at 5:52
• Let $y_0$ be the "first $y$" chosen in this process. Eventually we will need to construct the set $U_{y_0}$, which must contain $y_0$. Similarly, any $y \in U$ is an element of $U_y$ and therefore covered. – Austin Mohr May 17 '17 at 8:36
• Got it. Thank you. I wonder if it's possible to prove that $T_1$ is the weakest possible hypothesis that allows existence of these covers... – étale-cohomology May 18 '17 at 22:38
• One can push my argument through with something weaker (but I don't know if it has a name): For any open set $U$ and any point $x$ of $U$, there is an open neighborhood $V$ of $x$ that is a strict subset of $U$. This property is not as strong as $T_1$, since it still allows for the existence $x$ and $y$ that cannot be separated. However, you could take the $V$ in this property as the $U_x$ in my answer and it still get your desired refinement. This is property would be weakest possible, since otherwise the only way to cover $x$ is to use all of $U$. – Austin Mohr May 19 '17 at 1:01

It depends. Consider $\mathbb{R}$ with the usual topology and $U = (0,1)$, then we can write $(0,1) = \bigcup_{n=2}^\infty\left(\frac1n,1\right)$, for instance. Since every open set $U\subset\mathbb{R}$ is the union of intervals, it means that every open subset of $\mathbb{R}$ can be written as the union of proper open subsets of itself. It is easy to see that the same argument can be applied to general normed vector spaces.

However, if a singleton $\{x\}$ is an open set of your space then there are no proper open subsets of $\{x\}$ to unite.

Show that $K = [0,1]$ cannot be covered by open sets that are contained by $K$.

If $U$ is open, then $U$ can be covered by $\{U\}$. If $U$ is an open singleton, then there is no cover by proper subsets.

When $U$ is open, then for all $x$ in $U$, there is an open base set $U_x$ with $x$ in $U_x$ subset $U$. Often the base sets can be chosen to be proper subsets.