# What is the meaning of this definition

1. Dec 12, 2008

### transgalactic

lim inf (x_n) = inf {x: infinitely many x_n are < x }

i cant understand what they are saying here?

i was told to see lim inf as the infimum of all the limits of all the subsequences

they say something else
??

2. Dec 12, 2008

### HallsofIvy

Staff Emeritus
Look at all x with the property that there are infinitely many numbers in the sequence {x_n} less than x. Take the "infimum" (greatest lower bound).

For example, suppose the sequence is x_n= 1/n:
1, 1/2, 1/3, 1/4, 1/5, ...
If x is 0 or any negative number, all the numbers in the sequence are greater than x- an infinite number. If x is positive, 1/x is a positive number so (Archimedian property) there exist some integer N such that N> 1/x which means that 1/N< x. From that, in n> N, 1/n< 1/N< x so there are only a finite set of numbers in 1/n less than n. The set of all numbers, x, such that infinitely many members of the sequence are less than x is precisely the non-positive numbers, $(-\infty, 0]$. It's inf is 0, the limit of the sequence.

Or take a_n= -1/n for n odd, n/(n+1) for n even: -1, 2/3, -1/3, 4/5, -1/5, 6/7, -1/7... With an infinite number of negative numbers in there, it is easy to see that there exist an infinite number of terms of that sequence less than any non-negative number. But if x< 0, then -x is positive, so there exist N> -1/x or x> -1/N. There exist only a finite number of terms of the sequence less than x. Now the set of all x such that infinitely many x_n< x is the set of all negative numbers: $-\infty, 0)$. That set does not include 0 but includes number arbitrarily close to 0 so its infimum is still 0. Of course, the two "subsequential limits" are the limits of the subsequences {-1/n for n even} and {n/(n+1) for n odd} which are 0 and 1. The smaller of those, 0, is the lim inf.

3. Dec 13, 2008

### transgalactic

you are saying:

lim inf (x_n) = inf {x: infinitely many x_n are < x }

means that there are infinite number of members which are smaller then X

this is a definition of upper bound
so this X is a Upper bound

4. Dec 13, 2008

### mutton

No, an upper bound is greater than or equal to all $$x_n$$, not just infinitely many of them.

5. Dec 13, 2008

### HallsofIvy

Staff Emeritus
No I didn't say that and No that is NOT the definition of upper bound!
The definition of "X is an upper bound of S" is that ALL members of S are less than X, not just an infinite number of them.

6. Dec 13, 2008

### transgalactic

infinitely is not all
??