Let A be closed, non-empty subset of real numbers that has a lowerbound. Prove t

Question

Let A be closed, non-empty subset of real numbers that has a lowerbound. Prove that A contains its greatest lower bound.

Explanation / Answer

If A is closed then it contains all its limit points, say all thelimits of convergent sequences of elements of A. I will assumewithout loss of generality that A is an interval such that diam(A)= 2. I will make a remark later on how this is general enough. Because A has a lower bound, then there exists an M such that forevery a A, M 0 and g+1/n > g does not belong to A, so g isnot inf A, contrary to assumption. Then we have proved that everya_n A. Notice that the limit of {a_n} is g, so g A'.Because A is closed, g A. The remark: If the set A is connected in R, then it is an interval(it is a theorem). It is true that if A is closed, either g is anisolated point of A or it is a limit point of A. The trivial casehappens if g is not a limit point (is isolated), try to picture andit would make sense that our sequence(the one that converges to g)should be { g, g, g, ....} from some index. This is because A wouldhave the form of {g} U B, with B another closed set with everyelement of B greater than and distinct from g. So this case isactually not interesting for then g would actually be an element ofA, which is what we want! So I have to consider the casewhere the "lowest" portion of the set A is an interval. Here,to generalize the result assume that this "lower" part of A, callit now B, it is true that diam(B) = r. Then the sequence is definedby a_n = g + r/2n, the 1/2 multiplication just to ensure withoutquestion that a_1 belongs to B. You should notice it is notneeded.

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