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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.

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2 Answers 2

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$g$ 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

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  • $\begingroup$ Connectedness is not the crucial hypothesis, compactness is. $X$ and $Y$ can both be disconnected and the theorem still holds, provided $X$ is compact. $\endgroup$
    – RghtHndSd
    Commented Feb 3, 2019 at 14:31
  • $\begingroup$ @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 $\mathbb{R}$ $\endgroup$
    – user515010
    Commented Feb 3, 2019 at 14:33
  • $\begingroup$ @RghtHndSd Anyway, I edited the answer to make it clearer: is it okay now? $\endgroup$
    – user515010
    Commented Feb 3, 2019 at 14:34
  • $\begingroup$ Thank you very much, gabriele cassese for your answer. $\endgroup$
    – tchappy ha
    Commented Feb 3, 2019 at 14:40
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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.

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  • $\begingroup$ Thank you very much, RghtHndSd, for your answer. $\endgroup$
    – tchappy ha
    Commented Feb 3, 2019 at 14:40

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