Prove: let alpha be a lower bound of a subset E of an ordered set S. If alpha exist in E, then alpha= infE(greatest lower bound of E).
PF: Already have alpha is a lower bound of E. Let lambda be a different lower bound of E. If alpha precedes lambda, the lambda will not be a lower bound because alpha exist in E. So, lambda precedes alpha. Hence alpha=infE.
To prove that if alpha is a lower bound of a subset E of an ordered set S and alpha exists in E, then alpha must be the greatest lower bound (infimum) of E, we need to consider the definition of infimum.
The infimum of a set E in an ordered set S is a greatest lower bound if two conditions are met:
1. alpha is a lower bound of E, meaning for every element e in E, alpha is less than or equal to e.
2. For any other lower bound lambda of E, alpha is less than or equal to lambda.
Let's walk through the proof step by step:
1. Start with the given information that alpha is a lower bound of E.
- This means that for every element e in E, alpha is less than or equal to e.
2. Assume there exists another lower bound lambda of E.
- Let lambda be a different lower bound of E.
3. Consider the relationship between alpha and lambda.
- If alpha precedes lambda (alpha is less than lambda), then lambda cannot be a lower bound because alpha exists in E.
- In other words, if alpha is less than lambda, then there must be an element in E that is also less than lambda, contradicting the assumption that lambda is a lower bound of E.
4. Therefore, lambda must precede alpha.
- Since lambda precedes alpha, lambda is not greater than alpha, and it remains a lower bound of E.
5. Conclusion: alpha is the greatest lower bound of E (infimum).
- Since alpha is a lower bound of E and lambda is any other lower bound of E, and alpha precedes lambda, alpha satisfies the condition of being the greatest lower bound.
Hence, the proof shows that if alpha is a lower bound of a subset E of an ordered set S and alpha exists in E, then alpha is the greatest lower bound (infimum) of E.