Suppose $$X=\{1,2,3 \}$$ $$C=\{\{1\},\{2\},\{1,3\}\}$$ $$U=\{\{2\},\{1,3\}\}$$ Can I say that $C$ is a cover of $X$ and $U$ is a subcover of $C$? If not,why?
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$\begingroup$Always use the definitions. $\cup C$ is the set of the members of the members of $C.$ That is, $x\in \cup C\iff \exists d\in C\;(x\in d).$
$C$ is a cover of $X $ iff $\cup C\supset X.$ If $C$ is a cover of $X$ then $D$ is a sub-cover of $C$ iff $D\subset C$ and $\cup D\supset X.$
It would be more logical to say "$D$ is a sub($X$)-cover of $C$" or something like that, because whether or not $D$ is a sub-cover of $C$ depends upon the $X$ that is being covered, unless $\cup D=\cup C.$
With $X=\{1,2,3\}$ and $C=\{\{1\},\{2\},\{1,3\}\}$ we have
$$x\in \cup C\iff (x\in \{1\}\lor x\in \{2\}\lor x\in \{1,3\})\iff x\in \{1,2,3\}=X.$$ So $\cup C=X.$ So $C$ is a cover of $X.$
And $U=\{\{2\}\}, \{1,3\}\}\subset C$ with $\cup U=\cup C=X,\;$ so $ U$ is a sub-cover of $C.$
$\endgroup$ $\begingroup$Yes, you can say it, and it's even true ;): $\bigcup C = X$ and $\bigcup U = X$ with $U \subseteq C$.
Note that any cover of a finite space is finite (there are only finitely many subsets to make the cover) and is its own subcover.
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