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Unformatted text preview: Math508, Fall 2010 Jerry L. Kazdan Bonus Problem for Set 4
1. Deﬁne two real numbers x and y to be equal if x − y is an integer. We write x ≡ y (mod 1) .
Thus we have a “topological circle” whose “circumference” is one.
Let α be an irrational real number, 0 < α < 1 and consider its integer multiples, α , 2α , 3α . . .
(mod 1) . Show that this set is dense in 0 ≤ x ≤ 1 .
S OLUTION Given any ε > 0 we’ll show that every point in 0 ≤ x ≤ 1 is (mod 1) within ε of
an integer multiple of α .
Pick an integer K > 0 so that 1/K < ε . Partition the interval [0, 1] into the K intervals
[0, 1/K ], [1/K , 1/2K ], . . . , [(K − 1)/K , 1] , each of width 1/K < ε . Consider the K + 1 (distinct!) points α, 2α, . . . (K + 1)α , (mod 1) . Since there are K + 1 points and only K intervals,
at least 2 of them must lie in one of the intervals. Say jα and kα are in the same interval. Then
 jα − kα < 1/K < ε (mod 1) .
C ASE 1 k > j Let = k − j > 0 . Then 0 < α < ε (mod 1) . For some integer N a ﬁnite
number of the intervals
[0, α], [ α, 2 α], . . . [(N − 1) α, N α] (mod 1).
cover the interval [0, 1] . Thus every number in [0, 1] is within ε of one of the numbers α ,
. . . N α.
C ASE 2 k < j . This is essentially identical to C ASE 1
[Last revised: October 31, 2010] 1 ...
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This note was uploaded on 03/06/2012 for the course MATH 508 taught by Professor Staff during the Fall '10 term at UPenn.
 Fall '10
 STAFF
 Logic, Real Numbers

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