Question

Check wether the collection G, given by:

G’ = {]1/(n+2), 1/n[ : n∈N}

is an open cover of ]0,1[.

Answer 1

(0,1) has no finite subcover because,if possible it has a subcover

i.e then $U^n_{i=1}I_n \eqslantgtr (0,1)$

$I_n=(\frac{1}{n+2}, \frac{1}{n})$

But $\frac{1}{m+1}< \frac{1}{n}$ and $0<\frac{1}{m+1}<1$

But $\frac{1}{m+1} \notin U^n_{i=1}I_n$ So, $[I_n ]_{n=1}^m$ is not a subcover

That is to say for every m , we see that $\frac{1}{m+1} \notin U_{n=1}^m I_n$

So it has no subcover