Question

{F} check,whether the collection G,given by:

G'={]1/n+2,1/n[:n€N}

is an open cover of ]0,1[

Answer 1

If $f$  is continuous at $x_0$, then it should be the case that $lim_{x→x_0}f(x)=f(x_0).$ But we can argue that $f$ fails this condition when $x_0=1/n$ for any positive integer $n>1.$ Clearly there infinitely many of these points. For any positive integer $n>1$  there is a neighborhood $U=(1/(n+1),1/(n−1))$ of $1/n$ for which $f(x)=0$  if $x∈U$ and $x≠1/n$ . Therefore $lim_{x→1/n}f(x)=0≠1=f(1/n).$