1
$\begingroup$

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 $\{U_i\}_{i\in I}$ with

$U_i\subset X$ open,

$\bigcup_{i\in I}U_i=X$

and that $X$ is compact if from every open cover $\{U_i\}_{i\in I}$ finitely many $U_{i_1},...,U_{i_r}$ can be chosen such that $U_{i_1}\cup U_{i_2} \cup ... \cup U_{i_r}=X$.

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!

$\endgroup$
1
  • 1
    $\begingroup$ Your prof was not mistaken. It might help to recall the definition of the subspace topology $\endgroup$
    – jgon
    Mar 27 '19 at 16:56
4
$\begingroup$

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 $U = O \cap X$.

$\endgroup$
2
$\begingroup$

A subset $K$ of a metric space $X$ is compact if and only if for every any collection of open sets $U_i\subseteq X,i\in I$ for which $K\subseteq \bigcup_{i\in I} U_i$, there is a finite set $F\subseteq I$ for which $K\subseteq \bigcup_{i\in F}U_i$. 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 $K\times K$. It can be shown that $V\subseteq K$ is open in $(K,d|_K)$ if and only if $V=U\cap K$ for some open $U\subseteq X$.

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 $a,b\in \mathbb R$. Even though such sets are not open in $\mathbb R$, they are open when viewed as subsets of $K$ with the metric inherited from $\mathbb R$.

$\endgroup$
1
$\begingroup$

e.g. $[0,\frac12)\cap \mathbb{Q}$ is an open set of $X=[0,1]\cap \mathbb{Q}$ in its subspace topology.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.