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Sorry if the title is a bit unclear, but I'm stuck on the definition of compactness for metric spaces using open covers. So our professor wrote (word for word) that in a metric space $(X,d)$, an open cover is a family $U_i_iin I$ with

$U_isubset X$ open,

$bigcup_iin IU_i=X$

and that $X$ is compact if from every open cover $U_i_iin I$ finitely many $U_i_1,...,U_i_r$ can be chosen such that $U_i_1cup U_i_2 cup ... cup U_i_r=X$.

Is this definition correct? Because in one of our exercises, they ask us to find an open cover (that doesn't have a finite subcover) for $X=[0,1]cap Bbb Q$ with the Euclidian distance, but I can't for the life of me think of any set that is a subset of $X$, contains $0$ or $1$, and on top of that is an open set - those three things seem mutually exclusive to me. Was our prof mistaken in writing that $U_i$ is a subset of $X$ in the definition for open covers?

Any help would be greatly appreciated!

Open means relatively open. A subset of $U$ of $X = [0,1] cap mathbf Q$ is open if there is an open set $O subset mathbf R$ satisfying $U = O cap X$.

A subset $K$ of a metric space $X$ is compact if and only if for every any collection of open sets $U_isubseteq X,iin I$ for which $Ksubseteq bigcup_iin I U_i$, there is a finite set $Fsubseteq I$ for which $Ksubseteq bigcup_iin FU_i$. Note that $U_i$ need only be subsets of $X$.

However, $K$ being compact as a subset of $X$ is equivalent to $(K,d|_K)$ being a compact space. Here, $d|_K$ is the metric on $X$ with its domain restricted to $Ktimes K$. It can be shown that $Vsubseteq K$ is open in $(K,d|_K)$ if and only if $V=Ucap K$ for some open $Usubseteq X$.

For your problem, for example, letting $K=[0,1]cap mathbb Q$, then $(a,b)cap [0,1]cap mathbb Q$ would be an open set for any $a,bin mathbb R$. Even though such sets are not open in $mathbb R$, they are open when viewed as subsets of $K$ with the metric inherited from $mathbb R$.

e.g. $[0,frac12)cap mathbbQ$ is an open set of $X=[0,1]cap mathbbQ$ in its subspace topology.