When I was studying Munkres' Topology I noticed his definition of covering is:
"A collection $\mathcal{A}$ of subsets of a space $X$ is said to cover $X$, or to be a covering of $X$ if the union of elements of $\mathcal{A}$ equals to $X$."
I see a lot of definitions only require the union contains $X$. SO what confused me is that if we requires the union equals $X$, how could we find a covering of closed subspace $Y$ by open sets in $X$? (As mentioned in Lemma 26.1 the same book).
Lemma 26.1 Let $Y$ be a subspace of $X$. Then $Y$ is compact if and only if every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$.
As far as I know, any arbitrary union of open sets should be open. So does this theorem only applies to those sets being both open and closed?
Best Answer
Your definition of an open cover is correct. I think the issue is that there are two different notions of open cover here.
Indeed, going strictly by the definition of your open cover, it doesn't even make sense to have an open cover of a something which is not a topological space. So to have an open cover of a subset $Y$ of $X$, we need to understand $Y$ as a topological space. Indeed, we can endow $Y$ with a topology called the subspace topology. By definition, this is the set $\{U \cap Y : U \subseteq X \text{ is open}\}$. An open cover of $Y$ is therefore exactly that, an open cover under of this topological space with this given topology.
Alternatively, we can say that $\mathcal U$ covers the subset $Y \subseteq X$ if the union of $\mathcal U$ contains $Y$. This turns out to be basically the same as the above definition. Indeed, if $Y \subseteq \bigcup U_i$ ($U_i \subseteq X$ open), then $Y = \bigcup U_i \cap Y$, which are all open in the subspace topology we defined above. On the other hand, let $Y = \bigcup V_i$ be an open cover with all $V_i$ open in the subspace topology on $Y$. Then $V_i = Y \cap U_i$ for some $U_i \subseteq X$ open. Thus, $V_i \subseteq U_i$, so $Y \subseteq \bigcup U_i$, which is an open cover in this alternative sense.
The lemma you cited basically says that compactness can be defined by either of these notions of an open cover. That is, we have two notions of a compact subset, aligning to these two notions of open cover, and they are equivalent.
The first notion of compactness is that a space $Y$ is compact if any open cover $Y = \bigcup V_i$ admits a finite subcover. We therefore say that a subset $Y \subseteq X$ is compact if $Y$ is compact in the subspace topology. Alternatively, we say a subset $Y \subseteq X$ is compact if for any open cover $Y \subseteq \bigcup U_i$, there is a finite subcover $Y \subseteq U_{i_1} \cup \dots U_{i_k}$. The lemma says that $Y$ is compact in the first sense iff it is compact in the second sense.