Z każdego ograniczonego ciągu liczb rzeczywistych
(
c
n
)
n
=
0
∞
{\displaystyle (c_{n})_{n=0}^{\infty }}
rosnący ciąg indeksów
n
0
,
n
1
,
n
2
,
n
3
,
…
,
n
k
{\displaystyle n_{0},n_{1},n_{2},n_{3},\dots ,n_{k}}
(
c
n
k
)
k
=
0
∞
{\displaystyle (c_{n_{k}})_{k=0}^{\infty }}
Załóżmy, że
(
c
n
)
n
=
0
∞
{\displaystyle (c_{n})_{n=0}^{\infty }}
a
<
b
{\displaystyle a<b}
a
<
c
n
<
b
{\displaystyle a<c_{n}<b}
n
.
{\displaystyle n.}
a
k
,
b
k
∈
[
a
,
b
]
{\displaystyle a_{k},b_{k}\in [a,b]}
n
k
,
{\displaystyle n_{k},}
k
{\displaystyle k}
n
0
=
0
,
{\displaystyle n_{0}=0,}
a
0
=
a
,
{\displaystyle a_{0}=a,}
b
0
=
b
,
{\displaystyle b_{0}=b,}
n
k
<
n
k
+
1
,
{\displaystyle n_{k}<n_{k+1},}
a
k
⩽
a
k
+
1
⩽
c
n
k
+
1
⩽
b
k
+
1
⩽
b
k
,
{\displaystyle a_{k}\leqslant a_{k+1}\leqslant c_{n_{k+1}}\leqslant b_{k+1}\leqslant b_{k},}
b
k
−
a
k
=
(
b
−
a
)
⋅
2
−
k
,
{\displaystyle b_{k}-a_{k}=(b-a)\cdot 2^{-k},}
zbiór
{
n
:
c
n
∈
[
a
k
,
b
k
]
}
{\displaystyle \{n:c_{n}\in [a_{k},b_{k}]\}}
Pierwszy warunek powyżej definiuje
n
0
,
a
0
,
b
0
.
{\displaystyle n_{0},a_{0},b_{0}.}
n
k
,
a
k
,
b
k
{\displaystyle n_{k},a_{k},b_{k}}
d
=
a
k
+
b
k
2
.
{\displaystyle d={\frac {a_{k}+b_{k}}{2}}.}
{
n
:
c
n
∈
[
a
k
,
d
]
}
{\displaystyle \{n:c_{n}\in [a_{k},d]\}}
a
k
+
1
=
a
k
,
{\displaystyle a_{k+1}=a_{k},}
b
k
+
1
=
d
{\displaystyle b_{k+1}=d}
n
k
+
1
>
n
k
{\displaystyle n_{k+1}>n_{k}}
a
k
+
1
⩽
c
n
k
+
1
⩽
b
k
+
1
.
{\displaystyle a_{k+1}\leqslant c_{n_{k+1}}\leqslant b_{k+1}.}
{
n
:
c
n
∈
[
a
k
,
d
]
}
{\displaystyle \{n:c_{n}\in [a_{k},d]\}}
{
n
:
c
n
∈
[
d
,
b
k
]
}
{\displaystyle \{n:c_{n}\in [d,b_{k}]\}}
a
k
+
1
=
d
,
{\displaystyle a_{k+1}=d,}
b
k
+
1
=
b
k
{\displaystyle b_{k+1}=b_{k}}
n
k
+
1
>
n
k
{\displaystyle n_{k+1}>n_{k}}
a
k
+
1
⩽
c
n
k
+
1
⩽
b
k
+
1
.
{\displaystyle a_{k+1}\leqslant c_{n_{k+1}}\leqslant b_{k+1}.}
Po przeprowadzeniu powyższej konstrukcji zauważamy, że ciąg
(
c
n
k
)
k
=
0
∞
{\displaystyle (c_{n_{k}})_{k=0}^{\infty }}
ciągiem Cauchy’ego , a więc wobec zupełności prostej rzeczywistej jest on zbieżny.
Załóżmy, że
(
c
n
)
n
=
0
∞
{\displaystyle (c_{n})_{n=0}^{\infty }}
L
=
inf
{
c
n
:
n
}
,
{\displaystyle L=\inf\{c_{n}:n\},}
R
=
sup
{
c
n
:
n
}
,
{\displaystyle R=\sup\{c_{n}:n\},}
M
=
(
R
+
L
)
/
2
{\displaystyle M=(R+L)/2}
Δ
=
[
L
,
R
]
.
{\displaystyle \Delta =[L,R].}
Niech teraz
Δ
ϵ
=
[
L
ϵ
,
R
ϵ
]
{\displaystyle \Delta _{\epsilon }=[L_{\epsilon },R_{\epsilon }]}
Δ
{\displaystyle \Delta }
Δ
⟨
0
⟩
=
[
L
,
M
]
,
Δ
⟨
1
⟩
=
[
M
,
R
]
{\displaystyle \Delta _{\langle 0\rangle }=[L,M],\Delta _{\langle 1\rangle }=[M,R]}
Δ
ϵ
⟨
0
⟩
=
[
L
ϵ
,
M
ϵ
]
{\displaystyle \Delta _{\epsilon \langle 0\rangle }=[L_{\epsilon },M_{\epsilon }]}
Δ
ϵ
⟨
1
⟩
=
[
M
ϵ
,
R
ϵ
]
,
{\displaystyle \Delta _{\epsilon \langle 1\rangle }=[M_{\epsilon },R_{\epsilon }],}
gdzie
M
ϵ
=
(
L
ϵ
+
R
ϵ
)
/
2.
{\displaystyle M_{\epsilon }=(L_{\epsilon }+R_{\epsilon })/2.}
Łatwo zauważyć, że długość przedziału
Δ
ϵ
{\displaystyle \Delta _{\epsilon }}
(
R
−
L
)
/
2
n
,
{\displaystyle (R-L)/2^{n},}
n
{\displaystyle n}
ϵ
{\displaystyle \epsilon }
ϵ
′
,
{\displaystyle \epsilon ',}
ϵ
″
{\displaystyle \epsilon ''}
Δ
ϵ
″
⊆
Δ
ϵ
′
{\displaystyle \Delta _{\epsilon ''}\subseteq \Delta _{\epsilon '}}
wtedy i tylko wtedy, gdy ciąg
ϵ
′
{\displaystyle \epsilon '}
ϵ
″
.
{\displaystyle \epsilon ''.}
Łatwo też widać, że istnieje nieskończony rosnący ciąg indeksów
(
ϵ
n
)
n
,
{\displaystyle (\epsilon _{n})_{n},}
Δ
~
n
=
Δ
ϵ
n
,
{\displaystyle {\tilde {\Delta }}_{n}=\Delta _{\epsilon _{n}},}
n
∈
N
{\displaystyle n\in N}
(
c
n
)
n
.
{\displaystyle (c_{n})_{n}.}
Niech teraz
n
0
=
0
{\displaystyle n_{0}=0}
n
k
+
1
=
min
{
m
>
n
k
:
c
m
∈
Δ
~
k
}
.
{\displaystyle n_{k+1}=\min\{m>n_{k}:c_{m}\in {\tilde {\Delta }}_{k}\}.}
(
n
k
)
k
{\displaystyle (n_{k})_{k}}
c
n
k
∈
Δ
~
k
.
{\displaystyle c_{n_{k}}\in {\tilde {\Delta }}_{k}.}
Pokażemy, że ciąg
c
n
k
{\displaystyle c_{n_{k}}}
L
⋆
=
sup
{
L
~
n
:
n
∈
N
}
,
{\displaystyle L^{\star }=\sup\{{\tilde {L}}_{n}:n\in N\},}
L
~
n
=
L
ϵ
n
.
{\displaystyle {\tilde {L}}_{n}=L_{\epsilon _{n}}.}
Niech zatem
ε
>
0
{\displaystyle \varepsilon >0}
N
{\displaystyle N}
1
/
2
N
<
ε
/
2
{\displaystyle 1/2^{N}<\varepsilon /2}
K
{\displaystyle K}
L
⋆
−
ε
/
2
<
L
~
K
⩽
L
⋆
.
{\displaystyle L^{\star }-\varepsilon /2<{\tilde {L}}_{K}\leqslant L^{\star }.}
Biorąc teraz
k
⩾
max
{
N
,
K
}
{\displaystyle k\geqslant \max\{N,K\}}
|
c
n
k
−
L
⋆
|
<
|
c
n
k
−
L
~
k
|
+
|
L
~
k
−
L
⋆
|
=
|
c
n
k
−
L
~
k
|
+
(
L
⋆
−
L
~
k
)
⩽
1
/
2
k
+
(
L
⋆
−
L
~
K
)
<
ε
/
2
+
ε
/
2
=
ε
{\displaystyle |c_{n_{k}}-L^{\star }|<|c_{n_{k}}-{\tilde {L}}_{k}|+|{\tilde {L}}_{k}-L^{\star }|=|c_{n_{k}}-{\tilde {L}}_{k}|+(L^{\star }-{\tilde {L}}_{k})\leqslant 1/2^{k}+(L^{\star }-{\tilde {L}}_{K})<\varepsilon /2+\varepsilon /2=\varepsilon }
Tym samym wykazaliśmy zbieżność ciągu
(
c
n
k
)
k
{\displaystyle (c_{n_{k}})_{k}}
L
⋆
.
{\displaystyle L^{\star }.}
Niech
(
c
n
)
n
{\displaystyle (c_{n})_{n}}
L
=
inf
{
c
n
:
n
}
,
{\displaystyle L=\inf\{c_{n}:n\},}
R
=
sup
{
c
n
:
n
}
{\displaystyle R=\sup\{c_{n}:n\}}
d
=
(
R
−
L
)
/
2.
{\displaystyle d=(R-L)/2.}
Niech dalej
M
0
=
(
L
+
R
)
/
2
{\displaystyle M_{0}=(L+R)/2}
M
k
+
1
=
M
k
−
d
/
2
k
+
1
{\displaystyle M_{k+1}=M_{k}-d/2^{k+1}}
{
n
:
M
k
−
d
/
2
k
⩽
c
n
⩽
M
k
}
{\displaystyle \{n:M_{k}-d/2^{k}\leqslant c_{n}\leqslant M_{k}\}}
M
k
+
1
=
M
k
+
d
/
2
k
+
1
{\displaystyle M_{k+1}=M_{k}+d/2^{k+1}}
Δ
k
=
[
M
k
−
d
/
2
k
,
M
k
+
d
/
2
k
]
{\displaystyle \Delta _{k}=[M_{k}-d/2^{k},M_{k}+d/2^{k}]}
(
c
n
)
n
.
{\displaystyle (c_{n})_{n}.}
Ponieważ dla
k
=
0
,
{\displaystyle k=0,}
Δ
0
=
[
M
0
−
d
,
M
0
+
d
]
=
[
L
,
R
]
,
{\displaystyle \Delta _{0}=[M_{0}-d,M_{0}+d]=[L,R],}
Załóżmy zatem, że dla pewnego
k
{\displaystyle k}
Δ
k
=
[
M
k
−
d
/
2
k
,
M
k
+
d
/
2
k
]
{\displaystyle \Delta _{k}=[M_{k}-d/2^{k},M_{k}+d/2^{k}]}
(
c
n
)
n
.
{\displaystyle (c_{n})_{n}.}
{
n
:
M
k
−
d
/
2
k
⩽
c
n
⩽
M
k
}
{\displaystyle \{n:M_{k}-d/2^{k}\leqslant c_{n}\leqslant M_{k}\}}
M
k
+
1
=
M
k
−
d
/
2
k
+
1
{\displaystyle M_{k+1}=M_{k}-d/2^{k+1}}
Δ
k
+
1
=
[
M
k
+
1
−
d
/
2
k
+
1
,
M
k
+
1
+
d
/
2
k
+
1
]
=
[
M
k
−
d
/
2
k
,
M
k
]
,
{\displaystyle \Delta _{k+1}=[M_{k+1}-d/2^{k+1},M_{k+1}+d/2^{k+1}]=[M_{k}-d/2^{k},M_{k}],}
Jeśli zbiór
{
n
:
M
k
−
d
/
2
k
⩽
c
n
⩽
M
k
}
{\displaystyle \{n:M_{k}-d/2^{k}\leqslant c_{n}\leqslant M_{k}\}}
{
n
:
M
k
⩽
c
n
⩽
M
k
+
d
/
2
k
}
,
{\displaystyle \{n:M_{k}\leqslant c_{n}\leqslant M_{k}+d/2^{k}\},}
M
k
+
1
=
M
k
+
d
/
2
k
+
1
{\displaystyle M_{k+1}=M_{k}+d/2^{k+1}}
Δ
k
+
1
=
[
M
k
+
1
−
d
/
2
k
+
1
,
M
k
+
1
+
d
/
2
k
+
1
]
=
[
M
k
,
M
k
+
d
/
2
k
]
,
{\displaystyle \Delta _{k+1}=[M_{k+1}-d/2^{k+1},M_{k+1}+d/2^{k+1}]=[M_{k},M_{k}+d/2^{k}],}
Niech teraz
m
0
=
0
{\displaystyle m_{0}=0}
m
k
+
1
=
min
{
n
>
m
k
:
c
n
∈
Δ
k
+
1
}
.
{\displaystyle m_{k+1}=\min\{n>m_{k}:c_{n}\in \Delta _{k+1}\}.}
(
c
m
n
)
n
{\displaystyle (c_{m_{n}})_{n}}
(
c
n
)
n
.
{\displaystyle (c_{n})_{n}.}
L
n
=
M
n
−
d
/
2
n
{\displaystyle L_{n}=M_{n}-d/2^{n}}
L
⋆
.
{\displaystyle L^{\star }.}
L
⋆
=
lim
n
→
∞
c
m
n
.
{\displaystyle L^{\star }=\lim _{n\to \infty }c_{m_{n}}.}
Niech w tym celu
ε
>
0
{\displaystyle \varepsilon >0}
N
{\displaystyle N}
1
/
2
N
<
ε
/
2
{\displaystyle 1/2^{N}<\varepsilon /2}
K
{\displaystyle K}
L
⋆
−
ε
/
2
<
L
K
⩽
L
⋆
.
{\displaystyle L^{\star }-\varepsilon /2<{L}_{K}\leqslant L^{\star }.}
Biorąc teraz
n
⩾
max
{
N
,
K
}
{\displaystyle n\geqslant \max\{N,K\}}
|
c
m
n
−
L
⋆
|
<
|
c
m
n
−
L
n
|
+
|
L
n
−
L
⋆
|
=
(
c
m
n
−
L
n
)
+
(
L
⋆
−
L
n
)
⩽
1
/
2
n
+
(
L
⋆
−
L
K
)
<
ε
/
2
+
ε
/
2
=
ε
{\displaystyle |c_{m_{n}}-L^{\star }|<|c_{m_{n}}-{L}_{n}|+|{L}_{n}-L^{\star }|=(c_{m_{n}}-{L}_{n})+(L^{\star }-{L}_{n})\leqslant 1/2^{n}+(L^{\star }-{L}_{K})<\varepsilon /2+\varepsilon /2=\varepsilon }
Tym samym wykazaliśmy zbieżność ciągu
(
c
m
n
)
n
{\displaystyle (c_{m_{n}})_{n}}
L
⋆
.
{\displaystyle L^{\star }.}
Zauważmy, że
L
⋆
{\displaystyle L^{\star }}
(
M
n
)
n
{\displaystyle (M_{n})_{n}}
R
n
=
M
n
+
d
/
2
n
.
{\displaystyle R_{n}=M_{n}+d/2^{n}.}
Prawie bezpośrednio z twierdzenia Bolzana-Weierstrassa wynika twierdzenie Weierstrassa :
jeśli
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
funkcją ciągłą , to jej obraz jest zbiorem ograniczonym. Ponadto funkcja
f
{\displaystyle f}
c
,
d
∈
[
a
,
b
]
{\displaystyle c,d\in [a,b]}
∀
x
∈
[
a
,
b
]
f
(
d
)
⩽
f
(
x
)
⩽
f
(
c
)
.
{\displaystyle \forall x\in [a,b]\quad f(d)\leqslant f(x)\leqslant f(c).}
Pokażemy najpierw, że obraz funkcji
f
{\displaystyle f}
ograniczony . Ponieważ każdy z przedziałów funkcji
(
−
M
,
M
)
,
{\displaystyle (-M,M),}
M
∈
R
,
{\displaystyle M\in \mathbb {R} ,}
otwartym , a
f
{\displaystyle f}
ciągła , to otwarte (w zbiorze
[
a
,
b
]
{\displaystyle [a,b]}
f
−
1
[
(
−
M
,
M
)
]
.
{\displaystyle f^{-1}[(-M,M)].}
{
f
−
1
[
(
−
M
,
M
)
]
:
M
∈
R
}
{\displaystyle \{f^{-1}[(-M,M)]:M\in \mathbb {R} \}}
[
a
,
b
]
,
{\displaystyle [a,b],}
zwartości tego ostatniego, istnieją
M
1
,
…
,
M
s
>
0
,
{\displaystyle M_{1},\dots ,M_{s}>0,}
[
a
,
b
]
=
f
−
1
[
(
−
M
1
,
M
1
)
]
∪
…
∪
f
−
1
[
(
−
M
s
,
M
s
)
]
.
{\displaystyle [a,b]=f^{-1}[(-M_{1},M_{1})]\cup \ldots \cup f^{-1}[(-M_{s},M_{s})].}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
−
M
⩽
f
(
x
)
⩽
M
,
{\displaystyle -M\leqslant f(x)\leqslant M,}
M
=
max
{
M
1
,
…
,
M
S
}
,
{\displaystyle M=\max\{M_{1},\dots ,M_{S}\},}
f
{\displaystyle f}
Oznaczmy kres górny obrazu
f
{\displaystyle f}
d
~
.
{\displaystyle {\tilde {d}}.}
d
~
∈
R
{\displaystyle {\tilde {d}}\in \mathbb {R} }
(
d
n
)
n
=
0
∞
{\displaystyle (d_{n})_{n=0}^{\infty }}
[
a
,
b
]
{\displaystyle [a,b]}
f
(
d
n
)
{\displaystyle f(d_{n})}
d
~
.
{\displaystyle {\tilde {d}}.}
(
d
n
k
)
k
=
0
∞
{\displaystyle (d_{n_{k}})_{k=0}^{\infty }}
(
d
n
)
n
=
0
∞
{\displaystyle (d_{n})_{n=0}^{\infty }}
d
.
{\displaystyle d.}
f
{\displaystyle f}
f
(
d
)
=
lim
k
→
∞
f
(
d
n
k
)
=
d
~
.
{\displaystyle f({d})=\lim _{k\to \infty }f(d_{n_{k}})={\tilde {d}}.}
f
{\displaystyle f}
d
∈
[
a
,
b
]
{\displaystyle {d}\in [a,b]}
f
{\displaystyle f}
f
(
x
)
⩽
f
(
d
)
{\displaystyle f(x)\leqslant f(d)}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
W analogiczny sposób pokazujemy istnienie liczby
c
,
{\displaystyle c,}
f
(
c
)
=
inf
{
f
(
x
)
:
a
⩽
x
⩽
b
}
.
{\displaystyle f({c})=\inf\{f(x):a\leqslant x\leqslant b\}.}
Analiza założeń
edytuj
Założenie zwartości (domkniętości i ograniczoności) dziedziny funkcji (odcinka
[
a
,
b
]
{\displaystyle [a,b]}
f
:
(
0
,
1
]
∋
x
↦
1
/
x
∈
R
{\displaystyle f\colon (0,1]\ni x\mapsto 1/x\in \mathbb {R} }
f
:
R
∋
x
↦
e
x
∈
R
{\displaystyle f\colon \mathbb {R} \ni x\mapsto e^{x}\in \mathbb {R} }
Twierdzenie Weierstrassa jest używane m.in. w dowodzie twierdzenia Rolle’a [potrzebny przypis ] .
![{\displaystyle a_{k},b_{k}\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6564cae84a1c0c8a1a7f4736274f41ce36ee1755)
![{\displaystyle \{n:c_{n}\in [a_{k},b_{k}]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9db2efc63d1c96c8d69d9878c4e755a4435dc120)
![{\displaystyle \{n:c_{n}\in [a_{k},d]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d4c1d5a3fad32a508ca56eaa2773f5512a7b0e1d)
![{\displaystyle \{n:c_{n}\in [d,b_{k}]\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8830b3cf6c705a070a42a7178bb8449520faced5)
![{\displaystyle \Delta =[L,R].}](https://wikimedia.org/api/rest_v1/media/math/render/svg/923e847fa7bd10b0204af86da06080abb445d479)
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