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.

  • 2
    $\begingroup$ Theorems such as the one you have mentioned are immediate consequences of the completeness of real numbers. And you may have a look at various forms of completeness. $\endgroup$ – Paramanand Singh Jun 24 '18 at 3:57

I won't go into the philosophical discussions you are stimulating here. Instead I propose the following "direct sequential proof" that an upper semicontinuous function on an interval $[a,b]$ is bounded. It is essentially the same as yours, but avoids talking about nonexisting circumstances.

Consider the auxiliary function $$g(x):=e^{-f(x)}\qquad(a\leq x\leq b)\ ,$$ and define $$\alpha:=\inf_{a\leq x\leq b} g(x)\geq0\ .$$ Then for all $n\geq1$ there is an $x_n\in[a,b]$ with $g(x)<\alpha+{1\over n}$, and there is a subsequence $\bigl(x_{n_k}\bigr)_{k\geq1}$ with $$\lim_{k\to\infty}x_{n_k}=\xi\in[a,b]\ .$$ If $f(\xi)=:M$ then by assumption on $f$ there is a $\delta>0$ such that $$f(x)\leq M+1\qquad \forall\>x\in\>]\xi-\delta,\xi+\delta[\>\cap\>[a,b]\ .$$ It follows that for large enough $k$ we have $g\bigl(x_{n_k}\bigr)\geq e^{-M-1}$, so that $$\alpha=\lim_{k\to\infty}g\bigl(x_{n_k}\bigr)\geq e^{-M-1}>0\ .$$ This implies $$f(x)=\log {1\over g(x)}\leq\log{1\over\alpha}\qquad(a\leq x\leq b)\ .$$

  • $\begingroup$ See, I didn't know if I was asking a loose philosophical question, or a mathematically tractable one. I was hoping it was the latter. It's interesting to see that a direct proof is possible using sequences! $\endgroup$ – Ryan Jun 26 '18 at 13:48

Good question, and perhaps it is a bit beyond the scope of the questions that I should be answering (having only studied math for like 3 semesters in undergrad and forgetting it all for a time, much to my chagrin) but I will give it a shot...

For some claims, the sequential characterization might be the only useful characterization to our avail. For example, uniform convergence of shifted functions implies $f$ is uniformly continuous.

It turns out that using such sequential characterizations to constructively (so basically no law of excluded middle) prove a function $f:\mathbb{R} \to \mathbb{R}$ satisfies some continuity-related property on an interval $(a,b)$ or $[a,b]$ ($a<b$) is in general not going to be a very straightforward endeavor, if possible. If you are interested in furthering your understanding of these questions, then I would start with this paper about sequential continuity. As well as the referenced papers [6], [9], and [10]. Last I checked the latter 2 papers unfortunately require a jstor account, and so I actually have not viewed them myself.

Some additional examples of nonconstructive proofs involving sequential continuity:


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.