# min and max

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• Oct 6th 2007, 06:22 PM
CindyMichelle
min and max
Let (an) be a convergent sequence. Show that the set S = {an : n in N, the natural numbers} (the range of the sequence (an)) has a maximal or a minimal element.
If I assume that the sequence has no maximal or minimal element, then I know that both the inf(an) and the sup(an) are not in the set S. I don't see where the contradiction happens though. Could someone help me here?
• Oct 7th 2007, 01:07 AM
TheBrain
Assuming I understand the question, this is what I would do.

If (an) converges to a then that means that there exists an N in the Naturals such that if n is greater than N, |(an) - a| < d for some d > 0

Now there are a finite number of values more than d away from a, so the maximal element is either the maximum of these few values, or a+d. Similar for the minumum element.
• Oct 7th 2007, 06:04 AM
CindyMichelle
What about the sequence does it contradict though?
• Oct 7th 2007, 06:42 AM
Plato
Here is where a wee bit of topology is really useful in analysis.
You know that every bounded set has both an infimum and a supremum.
If either the inf or sup does not belong to the set, then there is a sequence of points from the set that converges to that number.

You are given that $\displaystyle \left( {a_n } \right)$ converges. Therefore, any subsequence from the range of $\displaystyle \left( {a_n } \right)$ also converges and has the same limit. Thus, if both the inf and sup do not belong to the range then we have two different limit points. Now there is your contradiction. So one of the inf or the sup belongs to the range. Say that it is the sup. Look again at the response from TheBrain. There would only a finite collection less that some number. Any finite set contains both its inf and its sup. Thus the range also contains its inf.