The standard definition of compactness of a set over the real line says that a set in the real line is compact iff every open covering has a finite subcovering.
Am wondering, is this definition also equivalent to the statement that a set in the real line is compact iff every covering by closed intervals has a finite subcovering ?
Note: Here, a closed interval $[a,b]$ is such that $a < b$ …. some texts however allow that $a=b$ but am not considering that case… why do most texts use open covering instead of the closed covering defined in this way ?
Best Answer
$[0,1]=[\frac 1 2, 1] \bigcup_{n\geq 3} [0,\frac 1 2 -\frac 1 n]$ and there is no finite subcover.
However, the converse is true. Suppose every cover of $K$ by closed intervals of positive length has finite subcover. Let $K \subset \bigcup (a_i,b_i)$. Then $K$ is covered by a finite number of the closed intervals $[a_i,b_i]$. Now each of the end points of these intervals that belong to $k$ will belong to one of the intervals in the original cover. Can you finish?