\fWe will show that sup(E) is contained in E {The proof for inf(E) will be similar to this proof}. As we know that an unbounded set cannot be compact (the cover Gn = (n, n) is sufficient to show this). This imply that E will have an upper bound, and hence will have a finite supremum, which we will call it as s = sup (E). Suppose let take the contradiction that s is such that it does belongs to E, and consider the cover Gn such that Gn = (, s - 1/ n). The union of all these sets mention above will cover from (, s), and we know that since s is an upper bound for E, it covers E. As we also know that no finite sub cover is sufficient to cover E. This is because any finite sub cover has a maximum index N , and so it will only covers ( , s 1 /N ) . If this sufficed to cover E, then (s 1/ N) would be an upper bound for E, and this contradict our previous assumption that s is the least upper bound and Thus it also contradict our assumption that S does belongs to E and E is not compact. HENCE it is proved that sup(E) is in E by contradiction approach