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 $(a,b)$?

  • $\begingroup$ 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 $f(a,b)$. $\endgroup$
    – xbh
    Nov 25, 2018 at 13:58
  • $\begingroup$ 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. $\endgroup$
    – xbh
    Nov 25, 2018 at 14:05
  • $\begingroup$ @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 $x\in[0,1)$ and $f(x)=x-1$ if $x\in[2,3]$, 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! $\endgroup$
    – Silent
    Nov 25, 2018 at 14:18
  • $\begingroup$ @Silent: but $f^{-1}$ is continuous on any compact subset of $f([0,1)\cup[2,3])$ and that is all you need. $\endgroup$ Nov 25, 2018 at 14:44

1 Answer 1


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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