And... let $x$ be $\inf\{X:[a,x]\text{ isn't compact}\}$
Let $x$ be $\inf\{X:[a,x]\text{ has no finite subcovering of }F\}$
$x$ is in $\{X:[a,x]\text{ has no finite subcovering of }F\}$ or not
(Note that the $\inf$ of the empty set is $\infty$... but I'll let it be $b$ here because we're just talking about subsets of $[a,b]$)
Let's say $x$ is in $\{X:[a,x]\text{ has no finite subcovering of }F\}$
So, for every $[a,y]$ where $y<x$, there is a finite subcovering
And, because $F$ is a covering, there is an open set containing $x$
and because it's open, there is a $y<x$ in $U$
So $[a,x]\subset[a,y]\cup U$
And $[a,y]$ has a finite subcovering
So just add $U$ to the finite subcovering
which contradicts where I said that $[a,x]$ has no finite subcovering
$x$ is not in $\{X:[a,x]\text{ has no finite subcovering of }F\}$
But $x=\inf\{X:[a,x]\text{ has no finite subcovering of }F\}$
So... $[a,x]$ has a finite subcovering. Call the open set containing $x$ $U$.
And $U$ contains a $y>x$.
So $[a,y]$ also has a finite subcovering, 'cause it's just the same subcovering as $[a,x]$
But this contradicts where I said that $x$ is the inf of that set. 'Cause that means every $[a,y]$ with $y>x$ has no finite subcovering
That means that $\{X:[a,x]\text{ has no finite subcovering of }F\}$ has to be the empty set
Which means that $[a,b]$ has a finite subcovering of $F$
And 'cause that works for every $F$, that means that $[a,b]$ is compact
I will be doing my master thesis on dictionary learning, and I am trying to understand basic concepts of the subject reading the following paper:
K-SVD: An Algorithm for Designing Overcomplete Dictionaries for Sparse Representation
In the following lines there are two conclusions or results...
