In the current Wikipedia entry for covers, it says that

if ${\displaystyle C=\lbrace U_{\alpha }:\alpha \in A\rbrace }$ is an indexed family of subsets ${\displaystyle U_{\alpha }\subset X}$ (indexed by the set ${\displaystyle A}$), then ${\displaystyle C}$ is a cover of ${\displaystyle X}$ if ${\displaystyle \bigcup _{\alpha \in A}U_{\alpha }\supseteq X}$.

Since a set cannot contain duplicates and all $U_\alpha$ are strict subsets of $X$, why does it say ${\displaystyle \bigcup _{\alpha \in A}U_{\alpha }\supseteq X}$ instead of ${\displaystyle \bigcup _{\alpha \in A}U_{\alpha } = X}$? What would be an example where $X$ is only a subset of its cover's union instead of being equal to it?


2 Answers 2


You are right, one could simply require that $\bigcup_{\alpha \in A} U_\alpha= X$. However, if you continue reading the Wikipedia article, you will find

Also, if $Y$ is a (topological) subspace of $X$, then a cover of $Y$ is a collection of subsets $C = \{U_\alpha \}_{\alpha \in A}$ of $X$ whose union contains $Y$, i.e., $C$ is a cover of $Y$ if $$ Y ⊆ ⋃_{α ∈ A} U_α .$$ That is, we may cover $Y$ with either sets in $Y$ itself or sets in the parent space $X$.

I think this is the reason for requiring $\bigcup_{\alpha \in A} U_\alpha \supseteq X$ - it keeps notation consistent in both situations.


Both are equivalent, as we always have $ \displaystyle\bigcup_{\alpha\in A}U_\alpha\subseteq A$. Nonetheless, one usually writes the condition as $\displaystyle\bigcup_{\alpha\in A}U_\alpha\supseteq A$, as that reinforces the idea of a cover: a collection of sets whose union encloses all of $A$.

  • $\begingroup$ In fact the question could rather be, "Why is the condition $U_\alpha \subset X$ included in the definition of "cover?". Does it serve any useful purpose? $\endgroup$ Mar 20 at 18:54
  • $\begingroup$ @MartinRattigan Yes, that is another, more interesting question. $\endgroup$ Mar 20 at 19:14

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