MHB Show that f is uniformly continuous

[evinda]

Hello!

I want to show that if $x_n',x_n'' \in A$ with $x_n'-x_n'' \to 0 \Rightarrow f(x_n'')-f(x_n'') \to 0$,then $f$ is uniformly continuous on $A$.

We suppose that $f$ is not uniformly continuous o $A$.
So, $ \exists \epsilon>0$ such that $\forall \delta>0$ and $ \forall y_n',y_n'' \in A$ with $|y_n'-y_n''|<\delta$ $\Rightarrow$ $|f(y_n')-f(y_n'')| \geq \epsilon$.
Is it right so far? And how can I continue? (Thinking)

 

[Opalg]

Hello!

I want to show that if $x_n',x_n'' \in A$ with $x_n'-x_n'' \to 0 \Rightarrow f(x_n'')-f(x_n'') \to 0$,then $f$ is uniformly continuous on $A$.

We suppose that $f$ is not uniformly continuous o $A$.
So, $ \exists \epsilon>0$ such that $\forall \delta>0$ and $ \forall y_n',y_n'' \in A$ with $|y_n'-y_n''|<\delta$ $\Rightarrow$ $|f(y_n')-f(y_n'')| \geq \epsilon$.
Is it right so far? And how can I continue? (Thinking)

When negating a statement containing quantifiers, each $\exists$ must be changed to $\forall$ and vice versa. The definition of uniform continuity states that $$ \forall \varepsilon>0\ \exists \delta>0\text{ such that } \forall x,x'\in A \text{ with } |x-x'|<\delta,\ |f(x) - f(x')| < \varepsilon.$$ The negation of that is that if $f$ is not uniformly continuous on $A$ then $$ \exists \varepsilon>0 \text{ such that }\forall \delta>0\ \exists x,x'\in A \text{ with } |x'-x''|<\delta,\ \text{ but }|f(x') - f(x'')| \geqslant \varepsilon.$$ Apply that statement with $\delta = 1/n$ ($n = 1,2,3,\ldots$). That will give you two sequences $x'_n$ and $x''_n$ that contradict the given property of $f$.

 

[evinda]

When negating a statement containing quantifiers, each $\exists$ must be changed to $\forall$ and vice versa. The definition of uniform continuity states that $$ \forall \varepsilon>0\ \exists \delta>0\text{ such that } \forall x,x'\in A \text{ with } |x-x'|<\delta,\ |f(x) - f(x')| < \varepsilon.$$ The negation of that is that if $f$ is not uniformly continuous on $A$ then $$ \exists \varepsilon>0 \text{ such that }\forall \delta>0\ \exists x,x'\in A \text{ with } |x'-x''|<\delta,\ \text{ but }|f(x') - f(x'')| \geqslant \varepsilon.$$ Apply that statement with $\delta = 1/n$ ($n = 1,2,3,\ldots$). That will give you two sequences $x'_n$ and $x''_n$ that contradict the given property of $f$.

Why do we take $\delta=\frac{1}{n}$ ? Because of the fact that $\frac{1}{n} \to 0$ ? (Wondering)

 

[Opalg]

Why do we take $\delta=\frac{1}{n}$ ? Because of the fact that $\frac{1}{n} \to 0$ ? (Wondering)

Yes. It is a convenient way to get two sequences $(x'_n)$ and $(x''_n)$ such that $x'_n-x''_n\to0.$

 

[evinda]

Yes. It is a convenient way to get two sequences $(x'_n)$ and $(x''_n)$ such that $x'_n-x''_n\to0.$

So,we suppose that $f$ is not uniformly continuous on $A$.So,
$ \exists \varepsilon>0 \text{ such that }\forall \delta>0\ \exists y_n,y_n'\in A \text{ with } |y_n-y_n'|<\delta,\ \text{ but }|f(y_n) - f(y_n')| \geqslant \varepsilon.$

We pick $\delta=\frac{1}{n} \to 0$,and because of the fact that $y_n,y_n'\in A$ and $ |y_n-y_n'|<\delta \to 0$,we conclude that $f(y_n)-f(y_n') \to 0$,that is a contradiction,as $|f(y_n) - f(y_n')| \geqslant \varepsilon.$Right?? (Thinking)

 

[evinda]

So,we suppose that $f$ is not uniformly continuous on $A$.So,
$ \exists \varepsilon>0 \text{ such that }\forall \delta>0\ \exists y_n,y_n'\in A \text{ with } |y_n-y_n'|<\delta,\ \text{ but }|f(y_n) - f(y_n')| \geqslant \varepsilon.$

We pick $\delta=\frac{1}{n} \to 0$,and because of the fact that $y_n,y_n'\in A$ and $ |y_n-y_n'|<\delta \to 0$,we conclude that $f(y_n)-f(y_n') \to 0$,that is a contradiction,as $|f(y_n) - f(y_n')| \geqslant \varepsilon.$Right?? (Thinking)

Or shouldn't I have used sequences? (Thinking)

 

[evinda]

Or is it better like that? (Thinking)

We suppose that $f$ is not uniformly continuous on $A$.So,
$ \exists \varepsilon>0 \text{ such that }\forall \delta>0\ \exists y,y'\in A \text{ with } |y-y'|<\delta,\ \text{ but }|f(y) - f(y')| \geqslant \varepsilon. $

We take $\delta_n=\frac{1}{n}$ and then $\forall y_n,y_n' \in A$ and $|y_n-y_n'|<\delta \Rightarrow y_n-y_n' \to 0$, we conclude that $f(y_n)-f(y_n') \to 0 $,that is a contradiction.