Open cover of a metric space

Question

I am trying to find a definition for the open cover of a metric space, but i cannot find it. So, if X is a metric space and A is a subset of X, then what is the definition for open cover of A? Can anyone help? Thank you.

Answer 1

An open cover of $A$ in $X$ is simply a family $\mathscr{U}$ of open sets in $X$ such that $A\subseteq\bigcup\mathscr{U}$. A relatively open cover of $A$ as a subspace of $X$ is a family $\mathscr{U}$ of open sets in $A$, i.e., of sets of the form $U\cap A$ for some open $U$ in $X$.

Example: Let $X=\Bbb R$, and let $A=(0,1)$. Then $$\mathscr{U}=\left\{\left(\frac1n,1\right):n\in\Bbb Z^+\right\}$$ is an open cover of $A$, because each $x\in A$ belongs to at least one member of $\mathscr{U}$. Specifically, if $x\in A$, then $x>0$, so there is a positive integer $n$ such that $\frac1n<x$, and $x\in\left(\frac1n,1\right)\in\mathscr{U}$.

Finally, a set $U\subseteq X$ is open if and only if it is a union of open balls: for each $x\in U$ there is an $\epsilon_x>0$ such that $B(x,\epsilon_x)\subseteq U$, where $B(x,\epsilon_x)=\{y\in X:d(x,y)<\epsilon_x\}$.

Answer 2

An open cover of a subset $A \subseteq X$, is a collection $\{U_i\}_{i \in I}$ of open sets in $X$, such that

$$A \subseteq \bigcup_{i \in I} U_i$$