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loismustdie
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#1
Report Thread starter 14 years ago
#1
I have NO idea how to do this question...

Let
h:R ->R be continuous and
k:R ->R be monotonic increasing.

Suppose that h(x) = k(x) for every rational number x.

Prove that h and k must be identically equal.

[can assume that for every real number y, there exists a sequence of irrational numbers (tn) tending to y as n->infinity and sequences of rational numbers (rn) and (sn) both tending to y as n->infinity with sn <= y <= rn]
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Jonny W
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#2
Report 14 years ago
#2
Fix any real number y.

Let (s_n) and (r_n) be rational sequences such that
(i) s_n <= y <= r_n for all n,
(ii) s_n -> y as n -> infinity,
(iii) r_n -> y as n -> infinity.

By (i) and the monotone increasing property of k,

k(s_n) <= k(y) <= k(r_n) for all n . . . . . (1)

Putting k(s_n) = h(s_n) and k(r_n) = h(r_n) into (1),

h(s_n) <= k(y) <= h(r_n) for all n . . . . . (2)

Since h is continuous at y, it follows from (ii) and (iii) that h(s_n) -> h(y) and h(r_n) -> h(y) as n -> infinity. So (2) says that k(y) is sandwiched between two functions of n, each of which tends to h(y) as n -> infinity. So k(y) = h(y).
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loismustdie
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#3
Report Thread starter 14 years ago
#3
Wow, that simple...
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Jonny W
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#4
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#4
Is the result still true if we replace the assumption that h is continuous with the assumption that it is monotone increasing? Give a proof or counterexample. Rep for the first correct answer!
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RichE
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#5
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#5
(Original post by Jonny W)
Is the result still true if we replace the assumption that h is continuous with the assumption that it is monotone increasing? Give a proof or counterexample. Rep for the first correct answer!
No - for example consider

h(x)
= {x if x < pi,
= {4 if x = pi
= {x+2 if x > pi

k(x)
= {x if x < pi,
= {5 if x = pi,
= {x+2 if x>pi

Then h(x) = k(x) except at the irrational pi
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