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I'm going through James Dugundji's Topology but I'm having trouble with chapter XI, section 4 problem 2:

  1. Let $\mathscr{F}$ be the family of all continuous maps $I\to I$ such that $|f(s)-f(t)|\leq |s-t|$. Define $$d^+(f,g)=\max_{0\leq t \leq 1}|f(t)-g(t)|.$$ Prove: $\mathscr{F}$ is compact.

Here $I=[0,1]$. I'm familiar with the main properties of compactness and know that $1°$ countable paracompact spaces are compact iff every sequence has convergent subsequence. I was trying to use this on $\mathscr{F}$ but had no success.

I'm looking for any hints on how to start this proof.

Edit: I'm attempting an elementary proof, Arzela-Ascoli comes after this chapter so it shouldn't be used. Function spaces and completeness hasn't been touched on yet (I guess completeness in $I$ should be available though).

Sebastián P. Pincheira
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    Perhaps Arzela-Ascoli might help here? – Mousedorff Mar 02 '23 at 17:12
  • That comes in the next chapter of the book. I'm trying to solve it only with the material presented prior to this problem: basically an elementary proof. – Sebastián P. Pincheira Mar 02 '23 at 17:17
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    @legionwhale do you mean the sequence of functions $f_n$ each defined by $x\mapsto n$? The thing is that $f_n$ ought be from $I$ to $I$ so this sequence is not on $\mathscr{F}$. – Sebastián P. Pincheira Mar 02 '23 at 18:30
  • @SebastiánP.Pincheira Ah, I missed that. My mistake. – legionwhale Mar 02 '23 at 19:09
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    Hmm maybe you shouldn't use Arzela-Ascoli directly but you could try to copy the argument; the argument itself is just a standard diagonalization. – Mousedorff Mar 02 '23 at 22:52
  • Your discussion of what you can use and cannot leaves me a little unclear about what is available with respect to conpleteness. Have you covered that a metric space is compact if and only if it is complete and totally bounded? Completeness can proven from completeness of the unit interval. And the contraction condition gives a fairly simple way of showing total-boundedness. In fact, this approach is probably a good way to prove the result even if you just rely directly on completeness on $I$ without reference to the function space. – Paul Sinclair Mar 03 '23 at 17:35

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