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.

 

[Nate810]

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.

 

[quicknote]

I can say that your approach seems reasonable and could potentially work. Using bisection to construct a Cauchy sequence to find the infimum of a set is a valid method. However, it is important to provide a more detailed explanation and proof to support your reasoning.

First, let us define the Cauchy sequence x_n as follows: x_1 is the midpoint of the interval [a, b] where a is the lower bound of E and b is any upper bound of E. Then, for n > 1, x_n is the midpoint of the interval [a_n, b_n], where a_n is the lower bound of E and b_n is the previous x_n.

Next, we can prove that this sequence is Cauchy. By construction, the intervals [a_n, b_n] are getting smaller and smaller, and since E is bounded from below, the sequence x_n is also bounded. Therefore, by the Nested Interval Theorem, the sequence x_n must converge to a point x.

Now, we need to show that x is the infimum of E. Since x_n is a Cauchy sequence, it must be that for any \epsilon > 0, there exists an N \in \mathbb{N} such that for all m, n > N, |x_m - x_n| < \epsilon. In other words, the distance between any two terms in the sequence becomes arbitrarily small as the terms approach the limit x.

Since x is the limit of x_n, we can choose \epsilon = \frac{1}{n} and find an N such that for all m > N, |x_m - x| < \frac{1}{n}. This means that x_n is within \frac{1}{n} of x for all n > N, and since x_n is an element of E, x must also be an element of E. Therefore, x is a lower bound of E.

Finally, we need to show that x is the greatest lower bound of E. Suppose there exists a lower bound y < x. This means that there exists an element y' \in E such that y' < y. However, since x_n is a Cauchy sequence that converges to x, for any \epsilon > 0, there exists an N \in \mathbb{N} such that for all n > N, |x_n