In Spivak's chapter on uniform convergence he asks to prove the following
THEOREM Let $\{f_n\}$ be sequence of continuous functions that converge pointwise to $0$ over $[a,b]$. If $0\leq f_{n+1}\leq f_n$ for each $x$ and $n$, then convergence is actually uniform over $[a,b]$.
Now, he asks, as he did in other occasions, to argue by contradiction and use Bolzano Weierstrass to find an "appropriate sequence $\{x_n\}$". I'm guessing he wants me to find a sequence that goes to $0$ but $f_n(x_n)\not \to 0$. I honestly didn't look at that option, but I wrote the following direct proof :
PROOF (This was awfully wrong)
I also don't see why it is essential that the $f_n$ are continuous. Could you provide a proof using Bolzano Weierstrass?
With this, it seems one can prove Dini's theorem, which seems an immediate result:
THEOREM Let $f_n\to f$ pointwise and monotonically over $[a,b]$, with each $f_n$ continuous, and $f$ continuous. Then $f_n\to f$ uniformly.
PROOF Assume $\{f_n\}$ increasing, and set $g_n=f-f_n$. Then the $g_n$ are continuous, $g_n\to 0$ pointwise and $$0\leq g_{n+1}\leq g_n$$
By the above, $g_n\to 0$ uniformly over $[a,b]$, that is, $f_n\to f $ uniformly over $[a,b]$. If $\{f_n\}$ is decreasing, consider $\{-f_n\}$.
Then Spivak asks
$(1)$ What if $f$ is not continuous?
$(2)$ What if we replace $[a,b]$ with $(a,b)$?
Best Answer
Assume that we can find $\delta>0$, a subsequence $\{f_{n_k}\}$ and a sequence $\{x_{n_k}\}$ such that $f_{n_k}(x_{n_k})\geqslant \delta$ for all $k$. Bolzano-Weierstrass theorem ($[a,b]$ is compact) allows us to extract of $\{x_{n_k}\}$ a subsequence, denoted $\{t_j\}$, converging to some $t$. We have for all integers $n$ and $m$, denoting $g_k$ the sequence indexed by the integers appearing in $t_k$, $$\delta\leqslant g_{m+n}(t_{m+n})\leqslant g_n(t_{m+n}).$$ Now we fix $n$, and take the $\limsup_{m\to +\infty}$. This gives for all integer $n$, $$\delta\leqslant\limsup_{k\to +\infty}g_k(t_k)\leqslant g_n(t),$$ a contradiction.
In the second theorem, take $a=0, b=1$.