I've been looking for simple problems online to improve my grasp on the basics of elementary analysis. I'm not sure how much context I should include to make this question understood, so I'll just include it all; my question will be at the very bottom of the post if the details aren't needed and you want to skip to it.
I found the following problem, and thought of two ways to solve it:
$\textbf{Problem}$: A real-valued function $f$ on a closed, bounded interval $[a,b]$ is said to be upper semicontinuous provided that for every $\epsilon>0$ and every $p\in[a,b]$, there is a $\delta=\delta(\epsilon,p)>0$ such if $x\in[a,b]$ and $|x-p|<\delta$ then $f(x)<f(p)+\epsilon$. Prove that an upper semicontinuous function is bounded above on $[a,b]$.
$\underline{\text{First Proof}}$: The argument is by contradiction. Suppose that $f$ is not bounded above. Then, for each $n\in\mathbb{N}$, there exists $x_n$ such that $f(x_n)>n$. Since the sequence $(x_n)$ is bounded, it contains a convergent subsequence, $(x_{n_k})$, converging to some $x^*\in[a,b]$. Given $\epsilon>0$, there exists $\delta>0$ such that for all $x$ satisfying $|x-x^*|<\delta$, we have $f(x)<f(x^*)+\epsilon$. There exists $K$ such that for all $k\geq K$ we have $|x_{n_k}-x^*|<\delta$, so that $f(x_{n_k})<f(x^*)+\epsilon$ for all $k\geq K$. This is a contradiction, since we must have $f(x_{n_k})>n_k\geq k$ for all $k$.
$\underline{\text{Second Proof}}$: Given an $\epsilon >0$, the set of open intervals $$\mathcal{O}=\big\{(p-\delta_{\epsilon,p}, p+\delta_{\epsilon,p}): p\in [a,b]\big\}$$ forms an open cover of $[a,b]$. Heine-Borel guarantees the existence of a finite subcover: $$[a,b] \subset \bigcup^{n}_{i=1}(p_i-\delta_{\epsilon,p_i}, p_i+\delta_{\epsilon, p_i}).$$ Let $f(p^*)=\max \{f(p_1),f(p_2),...,f(p_n)\}$. Then $f(x)<f(p^*)+\epsilon$ for all $x\in [a,b]$, so that $f(p^*)+\epsilon$ serves as an upper bound for $f$ on $[a,b]$.
After doing the second proof and rereading the first proof, I realized (assuming the proofs are correct): Both rely on the compactness of the interval $[a,b]$. The first proof uses the sequential characterization of compactness; the second uses the finite subcover definition of compactness.
I wondered: Can I use the sequential characterization to get a direct proof? I don't see how to do this, and doubt it can be done. So, here's my question: If a direct proof really $\textit{isn't}$ possible with the sequential characterization, is there some deeper reason for this? Does this suggest some underlying difference between the sequential characterization and the finite open subcover definition?
More generally, it's impossible not to notice that there are often several equivalent characterizations of a single concept in analysis (e.g., the $\epsilon$-$\delta$ vs. the sequential vs. the open set definitions of continuity), and sometimes one is more appropriate than another. If it happens that for a given problem or a given theorem, one characterization of a concept only allows for indirect proof whereas another characterization allows for a direct proof, is there always some reason for this? Does this tell us anything significant about the differences between the characterizations, or about what assumptions underlie the different characterizations?
I hope this question make sense. Please change the tag if another is more appropriate.
