There’s this theorem in my analysis book that I want to prove, which states that:
If
$k_{1} \subseteq k_{2} \subseteq k_{3} \subseteq ...$
Then the intersection
$\bigcap_{n=1}^{\infty}k_{n} \neq \varnothing$
My idea was to show that $k_{n}$ contains its supremum, since all the sets are compact and thus are bounded and closed:
$k_{1}$ is bounded $\implies$$k_{n}$ is bounded for all n $\in$ N$\implies$ sup($k_{n}$) exists for all n $\in$ N
$k_{n}$ is closed for all n $\in$ N $\implies$ sup($k_{n}$) $\in$$k_{n}$ for all n $\in$ N
Thus the intersection is non empty because every $k_{n}$ is non empty
Is this correct? If not, where did it go wrong?