Heine theorem without subsequences
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
118k639112
asked Dec 11 '18 at 14:55
T.D.
173113
add a comment |
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
118k639112
asked Dec 11 '18 at 14:55
T.D.
173113
add a comment |
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
118k639112
asked Dec 11 '18 at 14:55
T.D.
173113
I wanted to prove the following without the need of subsequences :
let $f$ be a continuous fonction on a closed interval of $mathbb{R}$ then it is uniformly continuous.
I found this proof but can't understand it
Let us assume every increasing sequence on [0;1] converges (if anyone can help me proving this) and the IVT
then let us create a sequence $u_n$
$begin{cases}u_0=0 \
u_{n+1}=1 text{ if } forall x>x_k, |f(x)-f(x_k)|<varepsilon text{ and we stop here} \
text{otherwise } u_{n+1}=min(x) text{ s.t } |f(x)-f(x_k)|=varepsilon\
end{cases}$
if this sequence is finite, let $eta=max(x_{n+1}-x_n)$ then $forall (x,y),forall varepsilon >0$
$$|x-y|leq eta implies |f(x)-f(y)|<2varepsilon $$
Otherwise we would have a contradiction.
Can anyone help me understanding
Thank you for your time.
proof-verification proof-writing proof-explanation
proof-verification proof-writing proof-explanation
edited Dec 11 '18 at 15:01
Bernard
118k639112
asked Dec 11 '18 at 14:55
T.D.
173113
edited Dec 11 '18 at 15:01
Bernard
118k639112
asked Dec 11 '18 at 14:55
T.D.
173113
edited Dec 11 '18 at 15:01
Bernard
118k639112
edited Dec 11 '18 at 15:01
Bernard
118k639112
edited Dec 11 '18 at 15:01
Bernard
118k639112
118k639112
asked Dec 11 '18 at 14:55
T.D.
173113
asked Dec 11 '18 at 14:55
T.D.
173113
asked Dec 11 '18 at 14:55
T.D.
173113
173113
add a comment |
add a comment |
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