Topologie de R Exercice 3. Exercice 6. un+1 − un > 0 Exercice 7. un

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R
ARa
A
A A ={x1, x2, . . . , xn, . . .}
xn−n→ ∞− > a
(sin(n)) [−1,1]
a∈R\QA={ma +n m ∈Z, n ∈N}AR
[−1,1] (sin n)
√m−√n
A={√m−√n m, n ∈N}R
A={n+p√2n, p ∈N, n +p√2>0, n2−2p2= 1}A
R+∗
a > 1 (xn)∀i6=j, |xi−xj| ≥ 1
|i−j|a
un+1 −un>0
(un)un+1 −un−n→ ∞− >0
(un)
un+1 −un>0
f: [0,1] −→ [0,1] u0∈[0,1] (un)f u0
un+1 −un−n→ ∞− >0 (un)
f
exp(iun)
(un) (exp(iun)) (|un+1 −un|)
α < π (un)
exp(iun)
(un)un+1 −un−n→ ∞− >0un−n→ ∞− >+∞
(exp(iun)) U
exp(iun)
(xn)u > 0, v > 0u
v/∈Q(eiuxn)
(eivxn) (xn)
un+p≤un+up
(un)∀n, p ∈N, un+p≤un+upun
n

R
∗
f:R−→ R1√2
f:R2−→ R2
X∈R2f(X) = f(X+ (1,0)) = f(X+ (0,1)) = f(AX)A=1 1
0 1

R
A\]a−1
n, a +1 n[
x < y a ∈Nn≥a⇒√n+ 1 −√n<y−x
b∈N√a−√b < x
c≥a x < √c−√b < y
n+p√2>1⇒n > 0, p > 0A∩]1,+∞[ 3 + 2√2
11
2a
1
4a
xkk→ →
f⇒(un)
`= lim inf un
nε > 0p(`−ε)p≤up≤(`+ε)p
n∈N0≤k < p uk+ (`−ε)np ≤unp+k≤uk+ (`+ε)np
GR
G G =RG f
1√2√2/∈QG
f f
y∈R\Qx7−→ f(x, y)
y∈Qf f (x, y)7−→ g(y)g1
f