Show that inverse of this continuous function is continuous












0














This question is from baby Rudin, Ch 5, exer 2:




Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.




The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'



Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.



So, how to show that $g$ indeed continuous on $(a,b)$?






real-analysis derivatives continuity







share|cite|improve this question






















  • Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
    – xbh
    Nov 25 at 13:58












  • As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
    – xbh
    Nov 25 at 14:05












  • @xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
    – Silent
    Nov 25 at 14:18










  • @Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
    – Matematleta
    Nov 25 at 14:44


















0














This question is from baby Rudin, Ch 5, exer 2:




Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.




The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'



Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.



So, how to show that $g$ indeed continuous on $(a,b)$?






real-analysis derivatives continuity







share|cite|improve this question






















  • Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
    – xbh
    Nov 25 at 13:58












  • As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
    – xbh
    Nov 25 at 14:05












  • @xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
    – Silent
    Nov 25 at 14:18










  • @Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
    – Matematleta
    Nov 25 at 14:44
















0












0








0


1





This question is from baby Rudin, Ch 5, exer 2:




Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.




The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'



Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.



So, how to show that $g$ indeed continuous on $(a,b)$?






real-analysis derivatives continuity







share|cite|improve this question













This question is from baby Rudin, Ch 5, exer 2:




Suppose $f'(x)>0$ in $(a,b)$. Hence $f$ is strictly increasing in $(a,b)$. Let $g=f^{-1}$. Prove $g'(f(x))=frac1{f'(x)}$.




The solution I have uses the fact thaty $g$, the inverse of $f$, is continuous. But Rudin, in Th 4.17 says that 'Suppose $f$ is continuous bijection from compact metric space $X$ to metric space $Y$, then $f^{-1}$ continuous from $Y$ to $X$.'



Since domain I have is not compact, I can't apply th 4.17 directly. So, I tried looking at $[c,d]subset(a,b)$ and noting $$g|_{f([c,d])}:f([c,d])to [c,d]$$ continuous by th 4.17. Now, I tried to apply pasting lemma, but realized that pasting lemma may not hold for infinitely many closed subsets of domain.



So, how to show that $g$ indeed continuous on $(a,b)$?






real-analysis derivatives continuity




real-analysis derivatives continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 25 at 13:45









Silent

2,64932050




2,64932050












  • Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
    – xbh
    Nov 25 at 13:58












  • As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
    – xbh
    Nov 25 at 14:05












  • @xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
    – Silent
    Nov 25 at 14:18










  • @Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
    – Matematleta
    Nov 25 at 14:44




















  • Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
    – xbh
    Nov 25 at 13:58












  • As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
    – xbh
    Nov 25 at 14:05












  • @xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
    – Silent
    Nov 25 at 14:18










  • @Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
    – Matematleta
    Nov 25 at 14:44


















Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58






Continuity is a local notion. We say $f$ is continuous on $(a,b)$ iff $f$ is continuous at each $x in (a,b)$. For your question, I think you do not need the pasting lemma. Your attempts actually completely showed the continuity of $g$ on $f(a,b)$.
– xbh
Nov 25 at 13:58














As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05






As an alternative, just prove the continuity by definition. Make use of the assumption that $f$ is continuous and strictly increasing. Maybe first prove the monotonicity of $g$, then the continuity.
– xbh
Nov 25 at 14:05














@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18




@xbh, thanks for reply. But, if we take function $f$ from $[0,1)cup[2,3]to[0,2]$ defined by $f(x)=x$ if $xin[0,1)$ and $f(x)=x-1$ if $xin[2,3]$, and consider $[c,d]subset[0,1)$ and $[c,d]subset[2,3]$, then $f^{-1}$ seems continuous by my above argument, but it is not!
– Silent
Nov 25 at 14:18












@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44






@Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)cup[2,3])$ and that is all you need.
– Matematleta
Nov 25 at 14:44












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It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.






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    It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.






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      It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.






      share|cite|improve this answer
























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        It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.






        share|cite|improve this answer












        It is sufficient to show that $f$ is an open map. So for any $(c,d)subset (a,b)$, observe that $f((c,d))=(f(c),f(d))$ since $f$ is a stricly increasing function. Hence $f$ maps any open set into an open set.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 25 at 16:03









        KnobbyWan

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