La continuïtat uniforme d'una funció s'expressa com:
∀
ε
>
0
,
∃
δ
>
0
∀
x
,
y
∈
M
|
(
d
M
(
x
,
y
)
<
δ
⇒
d
N
(
f
(
x
)
,
f
(
y
)
)
<
ε
)
,
{\displaystyle \forall \varepsilon >0,\ \exists \delta >0\ \forall x,y\in M\ |\left(d_{M}(x,y)<\delta \Rightarrow d_{N}(f(x),f(y))<\varepsilon \right),}
on
d
M
{\displaystyle d_{M}}
d
N
{\displaystyle d_{N}}
M
{\displaystyle M}
N
{\displaystyle N}
m im que
f
{\displaystyle f}
M
{\displaystyle M}
f ormement contínua , la negació de la continuïtat uniforme de
f
{\displaystyle f}
∃
ε
0
>
0
∀
δ
>
0
∃
x
,
y
∈
M
|
(
d
M
(
x
,
y
)
<
δ
∧
d
N
(
f
(
x
)
,
f
(
y
)
)
≥
ε
0
)
.
{\displaystyle \exists \varepsilon _{0}>0\ \forall \delta >0\ \exists x,y\in M\ |\ {\big (}d_{M}(x,y)<\delta \wedge d_{N}(f(x),f(y))\geq \varepsilon _{0}{\big )}.}
Triant
ε
0
{\displaystyle \varepsilon _{0}}
δ
{\displaystyle \delta }
i tiu tenim dos punts
x
{\displaystyle x}
y
{\displaystyle y}
M
{\displaystyle M}
Si triem
δ
=
1
n
{\displaystyle \delta ={\frac {1}{n}}}
n
=
1
,
2
,
3
,
.
.
.
{\displaystyle n=1,2,3,...}
successions
{
x
n
}
{\displaystyle \{x_{n}\}}
{
y
n
}
{\displaystyle \{y_{n}\}}
d
M
(
x
n
,
y
n
)
<
1
n
∧
d
N
(
f
(
x
n
)
,
f
(
y
n
)
)
≥
ε
0
.
{\displaystyle d_{M}(x_{n},y_{n})<{\frac {1}{n}}\wedge d_{N}(f(x_{n}),f(y_{n}))\geq \varepsilon _{0}.}
Com que
M
{\displaystyle M}
teorema de Bozen-Weierstrass demostra l'existència de dues subsucesiones convergents (
x
n
k
→
x
0
{\displaystyle x_{n_{k}}\rightarrow x_{0}}
y
n
k
→
y
0
{\displaystyle y_{n_{k}}\rightarrow y_{0}}
d
M
(
x
n
k
,
y
n
k
)
<
1
n
k
∧
d
N
(
f
(
x
n
k
)
,
f
(
y
n
k
)
)
≥
ε
0
.
{\displaystyle d_{M}(x_{n_{k}},y_{n_{k}})<{\frac {1}{n_{k}}}\wedge d_{N}(f(x_{n_{k}}),f(y_{n_{k}}))\geq \varepsilon _{0}.}
Definim ara la successió
{
|
x
n
k
−
y
n
k
|
}
=
{
d
M
(
x
n
k
,
y
n
k
)
}
≤
{
1
n
k
}
→
0
{\displaystyle \{|x_{n_{k}}-y_{n_{k}}|\}=\{d_{M}(x_{n_{k}},y_{n_{k}})\}\leq {\bigg \{}{\frac {1}{n_{k}}}{\bigg \}}\rightarrow 0}
Com que la successió
{
|
x
n
k
−
y
n
k
|
}
{\displaystyle \{|x_{n_{k}}-y_{n_{k}}|\}}
{
|
x
n
k
−
y
n
k
|
}
→
l
≤
0
⟹
l
=
0
{\displaystyle \{|x_{n_{k}}-y_{n_{k}}|\}\rightarrow l\leq 0\Longrightarrow l=0}
{
x
n
k
−
y
n
k
}
→
0
⟹
{
x
n
k
−
y
n
k
}
→
x
0
−
y
0
=
0
⟹
x
0
=
y
0
{\displaystyle \{x_{n_{k}}-y_{n_{k}}\}\rightarrow 0\Longrightarrow \{x_{n_{k}}-y_{n_{k}}\}\rightarrow x_{0}-y_{0}=0\Longrightarrow x_{0}=y_{0}}
Com que
f
{\displaystyle f}
x
0
{\displaystyle x_{0}}
{
f
(
x
n
k
)
}
→
f
(
x
0
)
{\displaystyle \{f(x_{n_{k}})\}\rightarrow f(x_{0})}
{
f
(
y
n
k
)
}
→
f
(
x
0
)
{\displaystyle \{f(y_{n_{k}})\}\rightarrow f(x_{0})}
{
f
(
x
n
k
)
−
f
(
y
n
k
)
}
→
0
{\displaystyle \{f(x_{n_{k}})-f(y_{n_{k}})\}\rightarrow 0}
d
N
(
f
(
x
n
k
)
,
f
(
y
n
k
)
)
≥
ε
0
{\displaystyle d_{N}(f(x_{n_{k}}),f(y_{n_{k}}))\geq \varepsilon _{0}}
La contradicció prova que la nostra suposició que
f
{\displaystyle f}
absurda : llavors
f
{\displaystyle f}
teorema .
Enllaços externs
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