# LUB - Suppose f is a function with domain [a, b]. Since we...

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Suppose f is a function with domain [ a,b ]. Since we will be concerend with subsets of [ a,b ], we will simplify notation by letting ( p,q ) · stand for ( p,q ) [ a,b ]. If X is a subset of [ a,b ], f ( X ) := { f ( x ) | x X } . Suppose c [ a,b ]. Recall what we mean when we say that f is continuous at c : For any real number ± > 0, there is a real number δ > 0 so that: for all x [ a,b ], | x - c | < δ implies | f ( x ) - f ( c ) | < ± . The condition above (oﬀset on its own line) is equivalent to: for all x ( c - δ,c + δ ) · , f ( x ) ( f ( c ) - ±,f ( c ) + ± ) , and to: f ( ( c - δ,c + δ ) · ) ( f ( c ) - ±,f ( c ) + ± ) . Fact. If X is a bounded subset of the real line and u is the least upper bound of X , then any open interval that contains u also contains points of X . Proof. If u were surrounded by an open interval containing no ponts of X , there would be points in that interval to the left of u that were upper bounds for X . Then u could not be the least upper bount. Boundedness Theorem.
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## This note was uploaded on 11/23/2011 for the course MATH 1551 taught by Professor Malisoff during the Fall '08 term at LSU.

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