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Show that the assumptions of the Heine-Borel theorem do not $\implies$ compactness in $C^0([a,b])$ ( Hint, consider the set of functions $S=\{f\in C^0([a,b]): f \ \text{is uniformly bounded by} \ 1\}$ ).

My attempt:

1.) $S$ is bounded: Being uniformly bounded means being bounded in the sup norm $\|\cdot\|$, so $S=\{f\in C^0([a,b]): \|f\|\leq 1\}$.

2.) $S$ is closed: Suppose $\{f_n\}\subset S$ converges to $f$. Since convergence in $\|\cdot\|$ is equivlent to uniform convergence, and the uniform limit of continuous function is continuous, $f\in C^0([a,b])$. Since a norm $\|\cdot\|$ on a vector space is continuous, $\|f\|=\|\lim_{n\to\infty}f_n\|=\lim_{n\to\infty}\|f_n\|\leq 1$. Hence $f\in S$, so $S$ is closed

3.) The sequence $\{f_n\}=x^n$ on $[0,1]$ converges pointwise to $f(x)= \begin{cases} 0 & 0 \leq x < 1 \\[2pt] 1, &\text{for $x=1$} \end{cases} $

Hence, any subsequence of $\{f_n\}$ that converges must converge to $f$. Since convergence here means uniform convergence, you get a contradiction because if $f_{n_k}\to f$ uniformly, then since each $f_{n_k}$ is continuous, so is $f$, but clearly $f$ is discontinuous at $x=1$. So $S$ is not (sequentially) compact.

Does this look ok?

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  • $\begingroup$ I think it is perfect $\endgroup$
    – guestDiego
    Jun 4, 2016 at 18:44
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    $\begingroup$ Don't write \text{lim}; just write \lim. The former does not yield proper spacing in things like $a\lim b$, and the latter affects the format in a "displayed" context, thus: $$ \lim_{x\to1} f(x), $$ coded as \lim_{x\to1} f(x). Also, notice that $||a||$ looks different from $\|a\|$ and if that's not conspicuous enough for you, notice that $||a|| ||b||$ looks different from $\|a\|\|b\|$. The latter is standard. I edited accordingly and also put in proper use of "cases" rather than an array, for a piecewise definition. $\qquad$ $\endgroup$ Jun 4, 2016 at 18:46
  • $\begingroup$ Also: "Heine" is the right spelling, and it may help to remember that it's pronounced as two syllables, the first rhyming with "sine" or "sign", and the last being pronounced like the "a" in "about". If that last vowel weren't there, it would be pronounced differently. $\qquad$ $\endgroup$ Jun 4, 2016 at 18:51
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    $\begingroup$ It's a good proof. Another way to show that the closed unit ball of $C[a,b]$ has an infinite closed discrete subspace is to take a pairwise-disjoint family $\{[a_n,b_n]: n\in N\}$ of intervals in $[a,b]$, with $a_n<b_n.$ Then take $f_n\in C[a,b]$ such that $ f_n(x)=0$ for $x\not \in [a_n,b_n]$, and $\|f_n\|=1.$ Then $\|f_m-f_n\|=1$ when $m\ne n.$....Remark : \|x\| gives $\|x\|$ , compared to ||x||, which gives $||x||$. $\endgroup$ Jun 4, 2016 at 21:16
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    $\begingroup$ Just another example of functions $f_n$ which make the same job (i.e. $\|f_n-f_m\|=1$): $$f_n(x)=\sin\left(2^n \pi\frac{x-a}{b-a}\right) $$ $\endgroup$
    – guestDiego
    Jun 4, 2016 at 22:13

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