Proving inf(x) = lim with Cauchy Sequence

[*melinda*]

Question:

Suppose there is a set E\subset \Re is bounded from below.
Let x=inf(E)
Prove there exists a sequence x_1, x_2,... \in E, such that x=lim(x_n).

I am not sure but it seems like my x=lim(x_n) =liminf(x_n).
In class we constructed a Cauchy sequence by bisection to find sup. To do this proof I was thinking that I should do the same, but do it to find inf.

Does this seem like it will work?
Any suggestions would be greatly appreciated.
Thanks.

 

[member 553083]

Yes, your approach is correct. You can use the same method of bisection to find the infimum of the set, and then construct a sequence that converges to it. The idea is to start with an interval [a,b] that covers E. Then, use the bisection method to find the infimum x. This means you will take the midpoint c of the interval [a,b] and check whether c is in E or not. If c is in E, then the infimum of E must be less than c, so you can then reduce the interval to [a,c]. If c is not in E, then the infimum of E must be greater than c, so you can reduce the interval to [c,b]. You can repeat this process until the interval is of size less than some arbitrary tolerance. Once you have found the infimum x, you can easily construct a sequence x_1, x_2, ... in E that converges to x by choosing elements of E that are close enough to x.