Asked by Ashley
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.
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.
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