$f$ is uniformly continuity

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Let $f:mathbbRto mathbbR$ be a bounded continuous function. Suppose for any $yin mathbbR$, $f^-1(y)$ is the empty set or a finite set. Then $f$ is uniformly continuous.

My idea :If $f$ is not uniformly continuous, there exists $a_n, b_n$ and $epsilon $ s.t., $|a_n-b_n|leq frac1n$ $|f(a_n)-f(b_n)|>epsilon$ . $f$ is bounded, so I can take convergence subsequence. But I don't know how to use cardinality of $f^-1(y)$

real-analysis continuity uniform-continuity

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asked Aug 6 at 15:22

B.T.O

251111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what if the two sequences run off to $infty$?
  – ncmathsadist
  Aug 6 at 15:28
  

  
  
  
  

add a comment | 

up vote
10
down vote

favorite

3

Let $f:mathbbRto mathbbR$ be a bounded continuous function. Suppose for any $yin mathbbR$, $f^-1(y)$ is the empty set or a finite set. Then $f$ is uniformly continuous.

My idea :If $f$ is not uniformly continuous, there exists $a_n, b_n$ and $epsilon $ s.t., $|a_n-b_n|leq frac1n$ $|f(a_n)-f(b_n)|>epsilon$ . $f$ is bounded, so I can take convergence subsequence. But I don't know how to use cardinality of $f^-1(y)$

real-analysis continuity uniform-continuity

share|cite|improve this question

asked Aug 6 at 15:22

B.T.O

251111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what if the two sequences run off to $infty$?
  – ncmathsadist
  Aug 6 at 15:28
  

  
  
  
  

add a comment | 

up vote
10
down vote

favorite

3

up vote
10
down vote

favorite

3

3

Let $f:mathbbRto mathbbR$ be a bounded continuous function. Suppose for any $yin mathbbR$, $f^-1(y)$ is the empty set or a finite set. Then $f$ is uniformly continuous.

My idea :If $f$ is not uniformly continuous, there exists $a_n, b_n$ and $epsilon $ s.t., $|a_n-b_n|leq frac1n$ $|f(a_n)-f(b_n)|>epsilon$ . $f$ is bounded, so I can take convergence subsequence. But I don't know how to use cardinality of $f^-1(y)$

real-analysis continuity uniform-continuity

share|cite|improve this question

asked Aug 6 at 15:22

B.T.O

251111

Let $f:mathbbRto mathbbR$ be a bounded continuous function. Suppose for any $yin mathbbR$, $f^-1(y)$ is the empty set or a finite set. Then $f$ is uniformly continuous.

My idea :If $f$ is not uniformly continuous, there exists $a_n, b_n$ and $epsilon $ s.t., $|a_n-b_n|leq frac1n$ $|f(a_n)-f(b_n)|>epsilon$ . $f$ is bounded, so I can take convergence subsequence. But I don't know how to use cardinality of $f^-1(y)$

real-analysis continuity uniform-continuity

share|cite|improve this question

asked Aug 6 at 15:22

B.T.O

251111

share|cite|improve this question

share|cite|improve this question

asked Aug 6 at 15:22

B.T.O

251111

asked Aug 6 at 15:22

B.T.O

251111

asked Aug 6 at 15:22

B.T.O

251111

251111

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what if the two sequences run off to $infty$?
  – ncmathsadist
  Aug 6 at 15:28
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  But what if the two sequences run off to $infty$?
  – ncmathsadist
  Aug 6 at 15:28
  

  
  
  
  

But what if the two sequences run off to $infty$?
– ncmathsadist
Aug 6 at 15:28

But what if the two sequences run off to $infty$?
– ncmathsadist
Aug 6 at 15:28

add a comment | 

1 Answer
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Take $epsilon>0$. Consider the sequence $f(1),f(2),...$ which has a limit point $C$ by Bolzano-Weierstrass. Since $f^-1(C)$ is finite/empty and $f$ is continuous, it cannot cross $C$ infinitely many times, so assume WLOG that $f(x)<C$ for all sufficiently large $x$. Similarly $f$ can't cross $C-epsilon$ infinitely many times, so for all large enough $x$ we have $f(x)in (C-epsilon,C)$ (as $C$ is a limit point, so there are infinitely many $f(x)$ in that range). Say this holds for $x>B$, which means $|f(x)-f(y)|<epsilon$ for $x,yin (B,infty)$.

We can do a similar thing to find $A$ such that $|f(x)-f(y)|<epsilon$ for $x,yin (-infty,A)$, and note that $[A-1,B+1]$ is compact, so $f$ is uniformly continuous on that interval, so we can find $delta<1$ such that $|f(x)-f(y)|<epsilon$ for $|x-y|<delta$. Therefore $f$ is uniformly continuous on $mathbb R$.

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answered Aug 6 at 15:51

Akababa

2,557922

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

Take $epsilon>0$. Consider the sequence $f(1),f(2),...$ which has a limit point $C$ by Bolzano-Weierstrass. Since $f^-1(C)$ is finite/empty and $f$ is continuous, it cannot cross $C$ infinitely many times, so assume WLOG that $f(x)<C$ for all sufficiently large $x$. Similarly $f$ can't cross $C-epsilon$ infinitely many times, so for all large enough $x$ we have $f(x)in (C-epsilon,C)$ (as $C$ is a limit point, so there are infinitely many $f(x)$ in that range). Say this holds for $x>B$, which means $|f(x)-f(y)|<epsilon$ for $x,yin (B,infty)$.

We can do a similar thing to find $A$ such that $|f(x)-f(y)|<epsilon$ for $x,yin (-infty,A)$, and note that $[A-1,B+1]$ is compact, so $f$ is uniformly continuous on that interval, so we can find $delta<1$ such that $|f(x)-f(y)|<epsilon$ for $|x-y|<delta$. Therefore $f$ is uniformly continuous on $mathbb R$.

share|cite|improve this answer

answered Aug 6 at 15:51

Akababa

2,557922

add a comment | 

up vote
4
down vote



accepted

Take $epsilon>0$. Consider the sequence $f(1),f(2),...$ which has a limit point $C$ by Bolzano-Weierstrass. Since $f^-1(C)$ is finite/empty and $f$ is continuous, it cannot cross $C$ infinitely many times, so assume WLOG that $f(x)<C$ for all sufficiently large $x$. Similarly $f$ can't cross $C-epsilon$ infinitely many times, so for all large enough $x$ we have $f(x)in (C-epsilon,C)$ (as $C$ is a limit point, so there are infinitely many $f(x)$ in that range). Say this holds for $x>B$, which means $|f(x)-f(y)|<epsilon$ for $x,yin (B,infty)$.

We can do a similar thing to find $A$ such that $|f(x)-f(y)|<epsilon$ for $x,yin (-infty,A)$, and note that $[A-1,B+1]$ is compact, so $f$ is uniformly continuous on that interval, so we can find $delta<1$ such that $|f(x)-f(y)|<epsilon$ for $|x-y|<delta$. Therefore $f$ is uniformly continuous on $mathbb R$.

share|cite|improve this answer

answered Aug 6 at 15:51

Akababa

2,557922

add a comment | 

up vote
4
down vote



accepted

up vote
4
down vote



accepted

Take $epsilon>0$. Consider the sequence $f(1),f(2),...$ which has a limit point $C$ by Bolzano-Weierstrass. Since $f^-1(C)$ is finite/empty and $f$ is continuous, it cannot cross $C$ infinitely many times, so assume WLOG that $f(x)<C$ for all sufficiently large $x$. Similarly $f$ can't cross $C-epsilon$ infinitely many times, so for all large enough $x$ we have $f(x)in (C-epsilon,C)$ (as $C$ is a limit point, so there are infinitely many $f(x)$ in that range). Say this holds for $x>B$, which means $|f(x)-f(y)|<epsilon$ for $x,yin (B,infty)$.

We can do a similar thing to find $A$ such that $|f(x)-f(y)|<epsilon$ for $x,yin (-infty,A)$, and note that $[A-1,B+1]$ is compact, so $f$ is uniformly continuous on that interval, so we can find $delta<1$ such that $|f(x)-f(y)|<epsilon$ for $|x-y|<delta$. Therefore $f$ is uniformly continuous on $mathbb R$.

share|cite|improve this answer

answered Aug 6 at 15:51

Akababa

2,557922

Take $epsilon>0$. Consider the sequence $f(1),f(2),...$ which has a limit point $C$ by Bolzano-Weierstrass. Since $f^-1(C)$ is finite/empty and $f$ is continuous, it cannot cross $C$ infinitely many times, so assume WLOG that $f(x)<C$ for all sufficiently large $x$. Similarly $f$ can't cross $C-epsilon$ infinitely many times, so for all large enough $x$ we have $f(x)in (C-epsilon,C)$ (as $C$ is a limit point, so there are infinitely many $f(x)$ in that range). Say this holds for $x>B$, which means $|f(x)-f(y)|<epsilon$ for $x,yin (B,infty)$.

We can do a similar thing to find $A$ such that $|f(x)-f(y)|<epsilon$ for $x,yin (-infty,A)$, and note that $[A-1,B+1]$ is compact, so $f$ is uniformly continuous on that interval, so we can find $delta<1$ such that $|f(x)-f(y)|<epsilon$ for $|x-y|<delta$. Therefore $f$ is uniformly continuous on $mathbb R$.

share|cite|improve this answer

answered Aug 6 at 15:51

Akababa

2,557922

share|cite|improve this answer

share|cite|improve this answer

answered Aug 6 at 15:51

Akababa

2,557922

answered Aug 6 at 15:51

Akababa

2,557922

answered Aug 6 at 15:51

Akababa

2,557922

2,557922

add a comment | 

add a comment | 

 

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