1. Apr 12, 2012

cragar

1. The problem statement, all variables and given/known data
{1,1/2,2/3,3/4,4/5......} is this set compact.
3. The attempt at a solution
I think this set is compact because it contains its cluster point which is 1.
is this correct?

2. Apr 12, 2012

micromass

Staff Emeritus
That is correct.

3. Apr 12, 2012

HallsofIvy

Staff Emeritus
You could prove that by appealing to "any closed and bounded subset of the real numbers is compact" or directly from the definition:
Let $\{U_\alpha\}$, for $\alpha[itex] in some index set, be an open cover for this set. Since 1 is in the set, 1 is in at least one of those- relabel the sets, if necessary, so that set is called [itex]U_1$. Since $U_1$ is open, there exist a point p, in that set, and a number $\delta> 0$ such that $N_\delta(p)= \{x | |x- p|<\delta\}$ is in $U_1$. It then follows that if $n> 1/\delta$ $|1- (n-1)/n|= 1/n< \delta$ so that all except a finite number of the fractions in the sequence are in $U_1$. Every number in the sequence, $(n-1)/n$ for n less than that might be in as separate set in the original sequence but there are only a finite number of them.