# About Theorem 4.17 on p.90 in "Principles of Mathematical Analysis 3rd Edition" by Walter Rudin.

On p.90 in "Principles of Mathematical Analysis 3rd Edition" by Walter Rudin.

Theorem 4.17
Suppose $$f$$ is a continuous 1-1 mapping of a compact metric space $$X$$ onto a metric space $$Y$$. Then the inverse mapping $$f^{-1}$$ defined on $$Y$$ by $$f^{-1}(f(x)) = x (x \in X)$$ is a continuous mapping of $$Y$$ onto $$X$$.

On p.93 in "Principles of Mathematical Analysis 3rd Edition" by Walter Rudin.

Example 4.21
Let $$X$$ be the half-open interval $$[0, 2 \pi)$$ on the real line, and let $$\mathbf{f}$$ be the mapping $$X$$ onto the circle $$Y$$ consisting of all points whose distance from the origin is $$1$$, given by $$\mathbf{f}(t) = (\cos t, \sin t) (0 \leq t < 2 \pi).$$ The continuity of the trigonometric functions cosine and sine, as well as their periodicity properties, will be established in Chap. 8. These results show that $$\mathbb{f}$$ is a continuous 1-1 mapping of $$X$$ onto $$Y$$.
However, the inverse mapping (which exists, since $$\mathbb{f}$$ is one-to-one and onto) fails to be continuous at the point $$(1, 0) = \mathbb{f}(0)$$.

On p.235 in "Calculus 4th Edition" by Michael Spivak.

Theorem 3
If $$f$$ is continuous and one-one on an interval, then $$f^{-1}$$ is also continuous.

Let $$X, Y$$ be metric spaces.
Let $$f : X \to Y$$ be a bijective mapping.
Theorem 4.17 says that if $$f$$ is continuous and $$X$$ is compact, then $$f^{-1}$$ is continuous.

Let $$X, Y$$ be metric spaces.
Let $$f : X \to Y$$ be a bijective mapping.
Example 4.21 says that if $$f$$ is continuous and $$X$$ is not compact, then $$f^{-1}$$ is not continuous in general.

Let $$I$$ be any interval.
Let $$f : I \to \mathbb{R}$$ be an injective function.
Then, $$g : I \ni x \mapsto f(x) \in f(I)$$ is a bijective function.
Theorem 3 says that if $$g$$ is continuous, then $$g^{-1}$$ is continuous.

This situation confuses me.

Is the above function $$g : I \to f(I)$$ so special?

By the way, I guess the following proposition is true:

Let $$S$$ be any subset of $$\mathbb{R}$$.
Let $$f : S \to \mathbb{R}$$ be an injective function.
Then, $$g : S \ni x \mapsto f(x) \in f(S)$$ is a bijective function.
If $$g$$ is continuous, then $$g^{-1}$$ is continuous.

$$g$$ is, indeed, quite special: a continous function maps a connected set into a connected set, and since the only connected sets in $$\mathbb{R}$$ are intervals, then $$g:I \to A$$ where $$A$$ is an interval. Thus, its image is an interval, and that's the "peculiarity" (that allows us to avoid compactness). With a more general $$Y$$ we can have much "stranger" connected sets than intervals, as the example from Rudin shows: there connectedness is not useful anymore and compactness becomes fundamental

• Connectedness is not the crucial hypothesis, compactness is. $X$ and $Y$ can both be disconnected and the theorem still holds, provided $X$ is compact. Commented Feb 3, 2019 at 14:31
• @RghtHndSd Yes, that's exactly what I meant: connectedness by itself is not sufficient, in general: it is in this particular case since we are in $\mathbb{R}$
– user515010
Commented Feb 3, 2019 at 14:33
• @RghtHndSd Anyway, I edited the answer to make it clearer: is it okay now?
– user515010
Commented Feb 3, 2019 at 14:34
• Thank you very much, gabriele cassese for your answer. Commented Feb 3, 2019 at 14:40

The domain $$[0, 2\pi)$$ is not compact, which is why the theorems don't apply.

Your guess proposition would be correct if you also include the hypothesis that $$S$$ is compact. Then it is essentially Rudin's 4.17.

• Thank you very much, RghtHndSd, for your answer. Commented Feb 3, 2019 at 14:40