TITLE: Exact Definition of an Open Covering for Topology
QUESTION [0 upvotes]: Definition 0: Let $Y\subseteq X$. Let $\lbrace A_i \rbrace_{i\in I}=\lbrace B_\alpha\subseteq X\mid\alpha\in J \rbrace$. Thus, $\lbrace A_i \rbrace_{i\in I}$ covers $Y$ iff $Y\subseteq \bigcup_{\alpha \in I}A_\alpha$.
Definition 1: Let $(X, T)$ be a topological space. Now, let $Y\subseteq X$. Let $T_1=\lbrace Y\cap U\mid U\in T \rbrace$. Thus, $T_1$ on $Y$ is induced by the topology $T$ on $X$ which yields a topological space $(Y, T_1)$.
Definition 2: Now, let $X$ be topological space. Let $Y\subseteq X$. A covering $\lbrace A_\alpha \rbrace_{\alpha \in I}$ of $Y$ is said to be an open covering of $Y$ iff $\forall \alpha \in I$, "$A_\alpha$ is an open subset of $X$".

Question: Saying "$A_\alpha$ is an open subset of $X$" is confusing me, because there are multiple possibilities where the open subsets of $X$ could come from. Instead of saying "$A_\alpha$ is an open subset of $X$" which option below $\textbf{(A.)}$, $\textbf{(B.)}$, or $\textbf{(C.)}$ is it?
$\textbf{(A.)}$ "$A_\alpha$ is an element in the topology $T_1$ on $Y$ where $T_1$ on $Y$ is induced
$\text{  }$ by the topology $T$ of $X$ (aka $A_\alpha\in T_1$)"
$\textbf{(B.)}$ "$A_\alpha$ is an element in the topology of $X$ (aka $A_\alpha\in T$)"
$\textbf{(C.)}$ "$A_\alpha$ is an either of these topologies (it doesn't matter)"

REPLY [1 votes]: An open covering of a subset $Y \subset X$ is in fact a family of open $A_\alpha \in T$ such that $Y \subset \bigcup_\alpha A_\alpha$. The subspace topology of $Y$ is irrelevant here. This means that (B.) is correct.
To see the difference between your conditions (A.) - (C.) let us say more precisely that $\mathfrak{A} = \lbrace A_\alpha \rbrace$ is an open covering of $Y$ in the space $X$.
The $A_\alpha \cap Y$ are open in the subspace topology $T_1$ on $Y$ and form an open covering of $Y$ in the space $Y$ which we denote by $\mathfrak{A} \cap Y$. 
Moreover, if you have any open covering $\mathfrak{B} = \lbrace B_\alpha \rbrace$ of $Y$ in the space $Y$ (which corrresponds to (A.)), then each $B_\alpha$ has the form $A_\alpha \cap Y$ with an open $A_\alpha \in T$. Therefore the $A_\alpha$ form an open covering $\mathfrak{A}$ of $Y$ in the space $X$ such that $\mathfrak{B} = \mathfrak{A} \cap Y$. Note that $\mathfrak{A}$ is in general not uniquely determined by $\mathfrak{B}$.
Condition (C.) is definitely never used.
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