Prove that any continuous integer-valued function of a real variable is constant.

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I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.

How do I use this to answer the question?
A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my
notes, in analysis books or anywhere online.

analysis

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asked Nov 12 '12 at 23:25

Denis

15126

add a comment | 

up vote
3
down vote

favorite

2

I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.

How do I use this to answer the question?
A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my
notes, in analysis books or anywhere online.

analysis

share|cite|improve this question

asked Nov 12 '12 at 23:25

Denis

15126

add a comment | 

up vote
3
down vote

favorite

2

up vote
3
down vote

favorite

2

2

I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.

How do I use this to answer the question?
A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my
notes, in analysis books or anywhere online.

analysis

share|cite|improve this question

asked Nov 12 '12 at 23:25

Denis

15126

I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.

How do I use this to answer the question?
A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my
notes, in analysis books or anywhere online.

analysis

share|cite|improve this question

asked Nov 12 '12 at 23:25

Denis

15126

share|cite|improve this question

share|cite|improve this question

asked Nov 12 '12 at 23:25

Denis

15126

asked Nov 12 '12 at 23:25

Denis

15126

asked Nov 12 '12 at 23:25

Denis

15126

15126

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
5
down vote



accepted

It is not necessarily true unless the domain is connected.

If the domain $X$ is connected and $f$ is continuous, then $f(X)$ is also connected. The only connected subsets of the integers are sets containing at most one point. Hence $f(X)$ (which is presumably non-empty) has exactly one point and $f$ is constant on $X$.

Another approach is to suppose that $y_1=f(x_1), y_2=f(x_2)$. Suppose $y_1leq y_2$ without loss of generality. Then by the intermediate value theorem, $f$ must take all the values in $[y_1,y_2]$. However, since the range is the integers, this can only be true if $f$ only takes one value.

Here's a third approach: Since $f$ is continuous, for each $x_0$ we can find a $delta>0$ such that if $|x-x_0|<delta$, then $|f(x)-f(x_0)| < frac12$. Since $f$ is integer valued, this means that $f(x) = f(x_0)$ for all $|x-x_0|<delta$.

Now let $t_+ = sup x $. If $t_+<infty$, then by continuity, we must have $f(t_+) = f(0)$, and by the previous paragraph, there is a $delta>0$ such that $f(x) = f(t_+)$ for all $x$ satisfying $|x-t_+| < delta$, which contradicts the definition of $t_+$. Hence $t_+ = infty$. Applying the same approach to $t_- = inf x $ shows that $t_- = -infty$, hence $f(x) =f(0)$ for all $x$.

share|cite|improve this answer

edited Nov 13 '12 at 0:30

answered Nov 12 '12 at 23:31

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
  – Denis
  Nov 12 '12 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
  – copper.hat
  Nov 12 '12 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! :) Much appreciated!
  – Denis
  Nov 12 '12 at 23:45
  

  
  
  
  

add a comment | 

up vote
2
down vote

I’m assuming that function of a real variable means a function defined on an interval, a ray, or the whole real line, as the result isn’t true otherwise. If you’re allowed to use familiar results from calculus, you can simply appeal to the intermediate value theorem: if $a<b$, $f(a)=m$, and $f(b)=nne m$, pick any non-integer $c$ between $m$ and $n$ and note that if $f$ is continuous, there must be some $xin(a,b)$ such that $f(x)=cnotinBbb Z$.

If you’re not allowed to use that, the proof is substantially harder, since in effect you have to prove a special case of that theorem.

share|cite|improve this answer

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

It is not necessarily true unless the domain is connected.

If the domain $X$ is connected and $f$ is continuous, then $f(X)$ is also connected. The only connected subsets of the integers are sets containing at most one point. Hence $f(X)$ (which is presumably non-empty) has exactly one point and $f$ is constant on $X$.

Another approach is to suppose that $y_1=f(x_1), y_2=f(x_2)$. Suppose $y_1leq y_2$ without loss of generality. Then by the intermediate value theorem, $f$ must take all the values in $[y_1,y_2]$. However, since the range is the integers, this can only be true if $f$ only takes one value.

Here's a third approach: Since $f$ is continuous, for each $x_0$ we can find a $delta>0$ such that if $|x-x_0|<delta$, then $|f(x)-f(x_0)| < frac12$. Since $f$ is integer valued, this means that $f(x) = f(x_0)$ for all $|x-x_0|<delta$.

Now let $t_+ = sup x $. If $t_+<infty$, then by continuity, we must have $f(t_+) = f(0)$, and by the previous paragraph, there is a $delta>0$ such that $f(x) = f(t_+)$ for all $x$ satisfying $|x-t_+| < delta$, which contradicts the definition of $t_+$. Hence $t_+ = infty$. Applying the same approach to $t_- = inf x $ shows that $t_- = -infty$, hence $f(x) =f(0)$ for all $x$.

share|cite|improve this answer

edited Nov 13 '12 at 0:30

answered Nov 12 '12 at 23:31

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
  – Denis
  Nov 12 '12 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
  – copper.hat
  Nov 12 '12 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! :) Much appreciated!
  – Denis
  Nov 12 '12 at 23:45
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

It is not necessarily true unless the domain is connected.

If the domain $X$ is connected and $f$ is continuous, then $f(X)$ is also connected. The only connected subsets of the integers are sets containing at most one point. Hence $f(X)$ (which is presumably non-empty) has exactly one point and $f$ is constant on $X$.

Another approach is to suppose that $y_1=f(x_1), y_2=f(x_2)$. Suppose $y_1leq y_2$ without loss of generality. Then by the intermediate value theorem, $f$ must take all the values in $[y_1,y_2]$. However, since the range is the integers, this can only be true if $f$ only takes one value.

Here's a third approach: Since $f$ is continuous, for each $x_0$ we can find a $delta>0$ such that if $|x-x_0|<delta$, then $|f(x)-f(x_0)| < frac12$. Since $f$ is integer valued, this means that $f(x) = f(x_0)$ for all $|x-x_0|<delta$.

Now let $t_+ = sup x $. If $t_+<infty$, then by continuity, we must have $f(t_+) = f(0)$, and by the previous paragraph, there is a $delta>0$ such that $f(x) = f(t_+)$ for all $x$ satisfying $|x-t_+| < delta$, which contradicts the definition of $t_+$. Hence $t_+ = infty$. Applying the same approach to $t_- = inf x $ shows that $t_- = -infty$, hence $f(x) =f(0)$ for all $x$.

share|cite|improve this answer

edited Nov 13 '12 at 0:30

answered Nov 12 '12 at 23:31

copper.hat

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
  – Denis
  Nov 12 '12 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
  – copper.hat
  Nov 12 '12 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! :) Much appreciated!
  – Denis
  Nov 12 '12 at 23:45
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

It is not necessarily true unless the domain is connected.

If the domain $X$ is connected and $f$ is continuous, then $f(X)$ is also connected. The only connected subsets of the integers are sets containing at most one point. Hence $f(X)$ (which is presumably non-empty) has exactly one point and $f$ is constant on $X$.

Another approach is to suppose that $y_1=f(x_1), y_2=f(x_2)$. Suppose $y_1leq y_2$ without loss of generality. Then by the intermediate value theorem, $f$ must take all the values in $[y_1,y_2]$. However, since the range is the integers, this can only be true if $f$ only takes one value.

Here's a third approach: Since $f$ is continuous, for each $x_0$ we can find a $delta>0$ such that if $|x-x_0|<delta$, then $|f(x)-f(x_0)| < frac12$. Since $f$ is integer valued, this means that $f(x) = f(x_0)$ for all $|x-x_0|<delta$.

Now let $t_+ = sup x $. If $t_+<infty$, then by continuity, we must have $f(t_+) = f(0)$, and by the previous paragraph, there is a $delta>0$ such that $f(x) = f(t_+)$ for all $x$ satisfying $|x-t_+| < delta$, which contradicts the definition of $t_+$. Hence $t_+ = infty$. Applying the same approach to $t_- = inf x $ shows that $t_- = -infty$, hence $f(x) =f(0)$ for all $x$.

share|cite|improve this answer

edited Nov 13 '12 at 0:30

answered Nov 12 '12 at 23:31

copper.hat

122k557156

It is not necessarily true unless the domain is connected.

If the domain $X$ is connected and $f$ is continuous, then $f(X)$ is also connected. The only connected subsets of the integers are sets containing at most one point. Hence $f(X)$ (which is presumably non-empty) has exactly one point and $f$ is constant on $X$.

Another approach is to suppose that $y_1=f(x_1), y_2=f(x_2)$. Suppose $y_1leq y_2$ without loss of generality. Then by the intermediate value theorem, $f$ must take all the values in $[y_1,y_2]$. However, since the range is the integers, this can only be true if $f$ only takes one value.

Here's a third approach: Since $f$ is continuous, for each $x_0$ we can find a $delta>0$ such that if $|x-x_0|<delta$, then $|f(x)-f(x_0)| < frac12$. Since $f$ is integer valued, this means that $f(x) = f(x_0)$ for all $|x-x_0|<delta$.

Now let $t_+ = sup x $. If $t_+<infty$, then by continuity, we must have $f(t_+) = f(0)$, and by the previous paragraph, there is a $delta>0$ such that $f(x) = f(t_+)$ for all $x$ satisfying $|x-t_+| < delta$, which contradicts the definition of $t_+$. Hence $t_+ = infty$. Applying the same approach to $t_- = inf x $ shows that $t_- = -infty$, hence $f(x) =f(0)$ for all $x$.

share|cite|improve this answer

edited Nov 13 '12 at 0:30

answered Nov 12 '12 at 23:31

copper.hat

122k557156

share|cite|improve this answer

share|cite|improve this answer

edited Nov 13 '12 at 0:30

edited Nov 13 '12 at 0:30

edited Nov 13 '12 at 0:30

answered Nov 12 '12 at 23:31

copper.hat

122k557156

answered Nov 12 '12 at 23:31

copper.hat

122k557156

answered Nov 12 '12 at 23:31

copper.hat

122k557156

122k557156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
  – Denis
  Nov 12 '12 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
  – copper.hat
  Nov 12 '12 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! :) Much appreciated!
  – Denis
  Nov 12 '12 at 23:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
  – Denis
  Nov 12 '12 at 23:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
  – copper.hat
  Nov 12 '12 at 23:42
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you! :) Much appreciated!
  – Denis
  Nov 12 '12 at 23:45
  

  
  
  
  

Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
– Denis
Nov 12 '12 at 23:40

Thanks for your answer! What do you mean by "connected domain"? Maybe I was taught a different term for it?
– Denis
Nov 12 '12 at 23:40

This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
– copper.hat
Nov 12 '12 at 23:42

This is what I mean en.wikipedia.org/wiki/Connected_space. I added an alternative approach using the intermediate value theorem.
– copper.hat
Nov 12 '12 at 23:42

Thank you! :) Much appreciated!
– Denis
Nov 12 '12 at 23:45

Thank you! :) Much appreciated!
– Denis
Nov 12 '12 at 23:45

add a comment | 

up vote
2
down vote

I’m assuming that function of a real variable means a function defined on an interval, a ray, or the whole real line, as the result isn’t true otherwise. If you’re allowed to use familiar results from calculus, you can simply appeal to the intermediate value theorem: if $a<b$, $f(a)=m$, and $f(b)=nne m$, pick any non-integer $c$ between $m$ and $n$ and note that if $f$ is continuous, there must be some $xin(a,b)$ such that $f(x)=cnotinBbb Z$.

If you’re not allowed to use that, the proof is substantially harder, since in effect you have to prove a special case of that theorem.

share|cite|improve this answer

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

add a comment | 

up vote
2
down vote

I’m assuming that function of a real variable means a function defined on an interval, a ray, or the whole real line, as the result isn’t true otherwise. If you’re allowed to use familiar results from calculus, you can simply appeal to the intermediate value theorem: if $a<b$, $f(a)=m$, and $f(b)=nne m$, pick any non-integer $c$ between $m$ and $n$ and note that if $f$ is continuous, there must be some $xin(a,b)$ such that $f(x)=cnotinBbb Z$.

If you’re not allowed to use that, the proof is substantially harder, since in effect you have to prove a special case of that theorem.

share|cite|improve this answer

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

add a comment | 

up vote
2
down vote

up vote
2
down vote

I’m assuming that function of a real variable means a function defined on an interval, a ray, or the whole real line, as the result isn’t true otherwise. If you’re allowed to use familiar results from calculus, you can simply appeal to the intermediate value theorem: if $a<b$, $f(a)=m$, and $f(b)=nne m$, pick any non-integer $c$ between $m$ and $n$ and note that if $f$ is continuous, there must be some $xin(a,b)$ such that $f(x)=cnotinBbb Z$.

If you’re not allowed to use that, the proof is substantially harder, since in effect you have to prove a special case of that theorem.

share|cite|improve this answer

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

I’m assuming that function of a real variable means a function defined on an interval, a ray, or the whole real line, as the result isn’t true otherwise. If you’re allowed to use familiar results from calculus, you can simply appeal to the intermediate value theorem: if $a<b$, $f(a)=m$, and $f(b)=nne m$, pick any non-integer $c$ between $m$ and $n$ and note that if $f$ is continuous, there must be some $xin(a,b)$ such that $f(x)=cnotinBbb Z$.

If you’re not allowed to use that, the proof is substantially harder, since in effect you have to prove a special case of that theorem.

share|cite|improve this answer

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

share|cite|improve this answer

share|cite|improve this answer

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

answered Nov 12 '12 at 23:44

Brian M. Scott

448k39492879

448k39492879

add a comment | 

add a comment | 

 

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