Does a cover have to be a subset of the topology? Say I have a topological space $M = (X, T)$ where $T = \{t_i \subset X\}$ is a collection of subsets of $X$.
Then if I define a cover $C = \{c_i \subset X\}$ such that $\bigcup_{i} c_i = X$, must the cover be a subset of the topology? That is, is it true that $C \subset T$? Or can the cover be completely unrelated to the topology?
 A: The general definition in use requires a covering to be just a set (or sequence) $\mathfrak U$ of (non-empty?) subsets of $X$ such that $\bigcup\mathfrak U=X$. The term is used mostly in topology, so it is natural that the cases of interest will be related to the topology. Open covers are more interesting for several reasons. Another interesting class of coverings I can think of is the class of fundamental coverings:

A fundamental covering of a topological space $(X,\tau)$ is a covering $\mathfrak U$ which satisfies the following equivalent conditions:

*

*$V\subseteq X$ is open if and only if, for all $U\in\mathfrak U$, $V$ is open in the subspace topology of $U$. In other words, if and only if for all $U\in\mathfrak U$ there is an open set $W\in\tau$ such that $V\cap U=U\cap W$.


*for any topological space $Y$, a function $f:X\to Y$ is continuous if and only if $\left.f\right\rvert_U$ is continuous (w.r.t. the subspace topology) for all $U\in\mathfrak U$.

A noticeable class of non-open coverings which satisfies (2) is the class of locally finite closed coverings, i.e. the closed coverings $\mathfrak U$ such that, for all $x\in X$, there exists a neighbourhood $V\ni x$ such that $C\cap V\ne\emptyset$ for a finite number of sets $C\in\mathfrak U$.
In a way, it generalises the property of the real line that says that, for instance, $$f=\begin{cases}3x+a&\text{if }x\le0\\-9x^2+b&\text{if }x<0\end{cases}$$
is continuous if and only if $a=b$.
