# Limit Inferior Proof

• Oct 25th 2009, 06:16 PM
utopiaNow
Limit Inferior Proof
Let $\{s_n\}_{n \in \mathbb{Z}^+}$ be a sequence of real numbers that is bounded below. Prove that $\displaystyle\liminf_{n\rightarrow\infty} s_n = \sup_m \left(\inf_{n > m} s_n\right)$.

I honestly can't even understand the question. I understand that if we let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{nk} \rightarrow x$ for some subsequence $\{s_{nk}\}$ then $\liminf_{n\rightarrow\infty} s_n = \inf{E}$ refers to the lower limit.

But I'm not really understanding the term on the RHS, much less how to go about proving the equality.

Any hints or suggestions would be greatly appreciated.
• Oct 25th 2009, 06:28 PM
rn443
Quote:

Originally Posted by utopiaNow
Let $\{s_n\}_{n \in \mathbb{Z}^+}$ be a sequence of real numbers that is bounded below. Prove that $\displaystyle\liminf_{n\rightarrow\infty} s_n = \sup_m \left(\inf_{n > m} s_n\right)$.

I honestly can't even understand the question. I understand that if we let $E$ be the set of numbers $x$ (in the extended real number system) such that $s_{nk} \rightarrow x$ for some subsequence $\{s_{nk}\}$ then $\liminf_{n\rightarrow\infty} s_n = \inf{E}$ refers to the lower limit.

But I'm not really understanding the term on the RHS, much less how to go about proving the equality.

Any hints or suggestions would be greatly appreciated.

Take a sequence a_n and define a new sequence b_n as b_n = inf{a_n, a_(n+1), a_(n+2), ...}. Then lim inf a_n is defined as lim b_n - that's the mathematical meaning of the term "lower limit." You're being asked to prove lim b_n = sup b_n. This can be done by proving b_n is nondecreasing.
• Oct 25th 2009, 06:46 PM
utopiaNow
Quote:

Originally Posted by rn443
Take a sequence a_n and define a new sequence b_n as b_n = inf{a_n, a_(n+1), a_(n+2), ...}. Then lim inf a_n is defined as lim b_n - that's the mathematical meaning of the term "lower limit." You're being asked to prove lim b_n = sup b_n. This can be done by proving b_n is nondecreasing.

Sorry if this is a naive question, but how does that define a sequence? Shouldn't the inf{a_n, a_(n+1), a_(n+2), ...} be just a single value?

Secondly won't showing that b_n is nondecreasing just show that lim b_n exists, won't I need to do extra work to show lim b_n = sup b_n?
• Oct 25th 2009, 06:55 PM
rn443
Quote:

Originally Posted by utopiaNow
Sorry if this is a naive question, but how does that define a sequence? Shouldn't the inf{a_n, a_(n+1), a_(n+2), ...} be just a single value?

It is a single value: the greatest lower bound of all the numbers a_n, a_(n+1), a_(n+2), ...

Quote:

Secondly won't showing that b_n is nondecreasing just show that lim b_n exists, won't I need to do extra work to show lim b_n = sup b_n?
The limit of a nondecreasing sequence, if it exists, is always its supremum. If we're dealing with the extended reals, we can drop the "if it exists" proviso.
• Oct 25th 2009, 07:05 PM
utopiaNow
Quote:

Originally Posted by rn443
It is a single value: the greatest lower bound of all the numbers a_n, a_(n+1), a_(n+2), ...

If it is a single value, doesn't that mean it's just a constant sequence and then won't proving that it's nondecreasing be unnecessary?
• Oct 25th 2009, 07:13 PM
rn443
Quote:

Originally Posted by utopiaNow
If it is a single value, doesn't that mean it's just a constant sequence and then won't proving that it's nondecreasing be unnecessary?

No. As you remove more and more terms, the greatest lower bound may change. For example, let a_n = n/(n+1). Then b_n = inf{n/(n+1), (n+1)/(n+2), (n+2)/(n+3), ...} = n/(n+1), which clearly changes each time you increase n.