text: # Trying to prove that if $f:[a, b]\to[s, t]$ is monotone then $f$ is continuous
  
  I'm trying to prove that if $f:[a, b]\to[s, t]$ is monotone (and its image is closed interval) then $f$ is continuous.
  
  My attempt:
  
  I say wlog, $f$ is increasing. I know that a monotone function only has jump discontinuities, so let's say $c\in [a, b]$ is a jump discontinuity, so:
  
  $$K=\lim\limits_{x\to c^-} f(x)<\lim\limits_{x\to c^+} f(x)=L$$
  
  So for each $c<x_n\to c$, $f(x_n)\to L$. $x_n>c$ so $f(x_n)\ge f(c)$ so $f(c)\le L$.
  
  And for each $c>y_n\to c$, $f(y_n)\to K$. $y_n<c$ so $f(y_n)\le f(c)$ so $f(c)\ge K$.
  
  So $L=K$ and this is the contradiction.
  
  And my question - where did I use that the image of $f$ is closed interval? Did I do something wrong?
  
  thanks?
  
  • Your conclusion that $L=K$ is not supported by what you have done so far. – Alex S Jul 7 '15 at 22:03
  • Oh I see.... Is this the right way though? – Stabilo Jul 7 '15 at 22:05
  • As you noticed, you need to incorporate the fact that the function's image is an interval. So there must be points in $[a,b]$ which map to points between $L$ and $K$, but how? – Alex S Jul 7 '15 at 22:07
  • Using my attempt of solution you mean, or another way? – Stabilo Jul 7 '15 at 22:09
  • After you conclude that $K\leq f(c)\leq L$, show that the existence of $K<y<f(c)$ or $f(c)<y<L$ leads to a contradiction. Conclude that $K=f(c)=L$. – Alex S Jul 7 '15 at 22:12
explanation: The text extract is a discussion about a proof attempt for a mathematical statement. The user is trying to prove that if a function \( f \) is monotone and maps a closed interval \([a, b]\) to another closed interval \([s, t]\), then \( f \) is continuous. The user presents their attempt at the proof and asks where they used the fact that the image of \( f \) is a closed interval. They also seek feedback on whether their approach is correct.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Trying to prove that if $f:[a, b]\to[s, t]$ is monotone then $f$ is continuous
  
  I'm trying to prove that if $f:[a, b]\to[s, t]$ is monotone (and its image is closed interval) then $f$ is continuous.
  
  My attempt:
  
  I say wlog, $f$ is increasing. I know that a monotone function only has jump discontinuities, so let's say $c\in [a, b]$ is a jump discontinuity, so:
  
  $$K=\lim\limits_{x\to c^-} f(x)<\lim\limits_{x\to c^+} f(x)=L$$
  
  So for each $c<x_n\to c$, $f(x_n)\to L$. $x_n>c$ so $f(x_n)\ge f(c)$ so $f(c)\le L$.
  
  And for each $c>y_n\to c$, $f(y_n)\to K$. $y_n<c$ so $f(y_n)\le f(c)$ so $f(c)\ge K$.
  
  So $L=K$ and this is the contradiction.
  
  And my question - where did I use that the image of $f$ is closed interval? Did I do something wrong?
  
  thanks?
  
  • Your conclusion that $L=K$ is not supported by what you have done so far. – Alex S Jul 7 '15 at 22:03
  • Oh I see.... Is this the right way though? – Stabilo Jul 7 '15 at 22:05
  • As you noticed, you need to incorporate the fact that the function's image is an interval. So there must be points in $[a,b]$ which map to points between $L$ and $K$, but how? – Alex S Jul 7 '15 at 22:07
  • Using my attempt of solution you mean, or another way? – Stabilo Jul 7 '15 at 22:09
  • After you conclude that $K\leq f(c)\leq L$, show that the existence of $K<y<f(c)$ or $f(c)<y<L$ leads to a contradiction. Conclude that $K=f(c)=L$. – Alex S Jul 7 '15 at 22:12
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The text extract is a discussion about a proof attempt for a mathematical statement. The user is trying to prove that if a function \( f \) is monotone and maps a closed interval \([a, b]\) to another closed interval \([s, t]\), then \( f \) is continuous. The user presents their attempt at the proof and asks where they used the fact that the image of \( f \) is a closed interval. They also seek feedback on whether their approach is correct.
  
  Label: question-and-answer