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I'm reading Intro to Topology by Mendelson.

The section is titled Connectedness of the Real Line, it's the section right before IVT is introduced.

I've been working on this problem, but I'm stuck on how to show that the function is onto. I know that because its monotone (strictly) increasing, that $f$ must be one-one. Intuitively, I know that the function must be onto, yet I'm having putting it down on paper.

To show the homeomorphism, I know that I have to show that either there is a continuous inverse function of $f$ or show that $f$ is a bijection and that every subset $O$ of $[a,b]$ is open if and only if $f(O)$ is open. To approach it using the second method I know that open sets in $\mathbb{R}$ are open intervals and for continuous functions the image of each interval is an interval, but do we know that an open interval maps to an open interval?

Thank you for any hints on how to approach this.

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    $\begingroup$ To prove that the map is surjective, simply use that the image of a connected set under a continuous map is again connected. There is no need to appeal to IVT directly. $\endgroup$ – Brandon Carter Jul 31 '13 at 4:20
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Surjectivity Hints:

  • Let $y \in \lbrack f(a),f(b) \rbrack$. Let $x_1 = \sup \{x \in \lbrack a,b \rbrack : f(x) < y\}$ and $x_2 = \inf\{x \in \lbrack a,b\rbrack : f(x) > y\}$. What's true about $x_1$ and $x_2$?

  • Using the answer you found to my first question and the continuity of $f$, can you prove that $f(x_1) = y$? (The easiest way to do this does not use IVT.)

  • This will prove that $\lbrack f(a),f(b) \rbrack \subset f(\lbrack a,b\rbrack)$. What about the opposite containment? (It's easier.)

Continuity Hints:

  • You now have a bijective function $f$ from $I = \lbrack a,b\rbrack$ onto another interval. Bijective $\implies$ $f^{-1}$ exists.
  • Let $\epsilon > 0$ and $a \leq x_0 \leq b$. The continuity of $f$ implies there exists $\delta > 0$ such that for all $x \in \lbrack a,b\rbrack$ such that $|x-x_0| < \delta$ we have $|f(x) - f(x_0)| < \epsilon$. Rewrite these inequalities in terms of $f^{-1}(y)$ and $f^{-1}(y_0)$ where $y = f(x)$ and $y_0 = f(x_0)$. Alternatively, note that this says $f((x_0-\delta,x_0+\delta)) \subset (f(x_0)-\epsilon,f(x_0)+\epsilon)$ and reach a conclusion from that.
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  • $\begingroup$ I'm short on time right now and won't be able to go through your response thoroughly. I'll work on this later and see where it goes. Thanks for the help! $\endgroup$ – Shant Danielian Jul 31 '13 at 4:14
  • $\begingroup$ Sure thing! Let me know if you have questions. $\endgroup$ – Dan Jul 31 '13 at 4:21
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Since $f$ is increasing, it is injective (proof?). One should actually prove, in my opinion, that $f([a,b])=[f(a),f(b)]$. It is clear that $f([a,b])\subseteq [f(a),f(b)]$ since $f$ is monotone. Can you prove the other inclusion? You do need the intermediate value theorem here! With this out of the way, one has a one one, onto function of the form $$f:[a,b]\to[c,d]$$

It's inverse then exists. Can you show it is continuous?

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  • $\begingroup$ To show the other inclusion, would you let $c\in[f(a),f(b)]$, then there exists some $d\in[a,b]$ such that $f(d)=c$ and so that would require $f(d)\in f([a,b])$. (I think I might be using IVT again) If I must use the intermediate value theorem, why is this problem right before the section, could it just be an error? $\endgroup$ – Shant Danielian Jul 31 '13 at 4:02
  • $\begingroup$ @ShantDanielian The proof is OK. Isn't it right after the section? $\endgroup$ – Pedro Tamaroff Jul 31 '13 at 4:06
  • $\begingroup$ Yeah, IVT is covered right after this section. $\endgroup$ – Shant Danielian Jul 31 '13 at 4:08
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From increasing you can easily show that $f$ is injective.
Let's show that $f:[a,b]\rightarrow\left[f(a),f(b)\right]$ is onto. If it's not onto and $S=\{x\in[a,b]:[f(a),f(x)]\subseteq[f(a),f(b)]\}, T=\{x\in[a,b]:[f(x),f(b)]\subseteq[f(a),f(b)]\}$ then by considering the $\sup S$ and $\inf T$ and taking the limits of $f(x)$ as $x\to\sup S$ and $x\to\inf T$ you can derive a contradiction.

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