Proof of uniform continuity of a continuous function on an interval [a,b] without using compactness arguments

Question

I give a proof of mine of the fact that if $f:[a,b]\rightarrow\mathbb{R}$ is a continuous function, then it is uniformly continuous. The idea is to avoid using compactness arguments (with coverings), but prove it in a direct way.

Proof: Let $\epsilon>0$, which is kept fixed. For each $x\in [a,b]$ we define the set:

$$A(x)=\{\delta\in(0,1] : \lvert y-x\rvert<\delta \Rightarrow\lvert f(x)-f(y)\rvert<\epsilon\}.$$

Every $A(x)$ is non-empty, because $f$ is continuous at $x$ and is clearly bounded above by $1$.

Therefore $\sup A(x)$ exists for all $x\in [a,b]$. We label

$$E(x)=\sup A(x),$$ $$\Delta=\inf\{E(x) : x\in [a,b]\}.$$

Our goal is to show that $\Delta>0$.

There is obviously a sequence such that $x_{i} \in [a,b]$ and such that $E(x_{i})\rightarrow\Delta$. There is also a subsequence (we don't change indices) such that $x_{i}\rightarrow\bar{x}\in[a,b]$. Let $\bar{\delta}=\sup A(\bar{x})$.

We define for each $c>0$ such that $0<c<\epsilon$, the set:

$$A(c)=\{\delta\in(0,1]:\lvert y-\bar{x}\rvert<\delta\Rightarrow\lvert f(y)-f(\bar{x})\rvert<\epsilon-c\}$$

and $\delta(c)=\sup A(c)$.

We will show that ${\displaystyle \lim_{c \to 0^{+}}}\delta(c)=\bar{\delta}$.

Indeed, we first observe that $\delta(c)\leq\bar{\delta}$ for every $c$ with $0<c<\epsilon$. (Because $\lvert f(y)-f(\bar{x})\rvert<\epsilon-c<\epsilon$).

Therefore $A(c)\subseteq A(\bar{x})$. Let $c_{n}\to 0^{+}$. Then, if for each $q<\bar{\delta}$ there only finite $n$ such that $\delta(c_{n})<q<\bar{\delta}$ then we have the definition of $\delta(c)\rightarrow\bar{\delta}$.

Now assume there is an infinity of $n$ for which $\delta(c_{n})<q<\bar{\delta}$.

Then for each $y$ with $\lvert y-\bar{x}\rvert <q<\delta$ we have $\lvert f(y)-f(x)\rvert<\epsilon$.

Now if there is one $n$ of this infinity of $n$ such that, for $y$ with $\lvert y-\bar{x}\rvert<q$ we have $\lvert f(y)-f(\bar{x})\rvert<\epsilon-c_{n}$, then $q\in A(c_{n})$ hence $q\leq\delta(c_{n})$, contradiction! Thus for each $n$ of the infinity there is an $y_{n}$ with $\lvert y_{n}-\bar{x}\rvert<q$ and $\lvert f(y_{n})-f(\bar{x})\rvert\geq\epsilon-c_{n}$'

Taking a subsequence (we don't change indices) $y_{n}\rightarrow y_{0}$ we get $\lvert y_{0}-\bar{x}\rvert\leq q$ and $\lvert f(y_{0})-f(\bar{x})\rvert\geq\epsilon$.

But $\lvert y_{0}-\bar{x}\rvert\leq q<\bar{\delta} $ implies $\lvert f(y_{0})-f(\bar{x})\rvert<\epsilon$, contradiction.

Therefore, $\delta(c_{n})\rightarrow\bar{\delta}$ and clearly there is a $c_{0}\in(0,\epsilon)$ with $\delta(c_{0})>\dfrac{\bar{\delta}}{2}$.

For the sequence $x_{i}$ and for large $i$'s (since $x_{i}\rightarrow\bar{x}$) we get $\lvert x_{i}-\bar{x}\rvert<\dfrac{\bar{\delta}}{2}$ and $\lvert f(x_{i})-f(\bar{x})\rvert<c_{0}$.

Let $y$ be such that $\lvert y-x_{i}\rvert<\delta(c_{0})-\dfrac{\bar{\delta}}{2}$.

Then $\lvert y-x_{i}\rvert+\lvert x_{i}-\bar{x}\rvert<\delta(c_{0})$ hence $\lvert y-\bar{x}\rvert<\delta(c_{0})$ which implies $\lvert f(y)-f(\bar{x})\rvert<\epsilon-c_{0}$. But $\lvert f(x_{i})-f(\bar{x})\rvert<c_{0}$

and hence $\lvert f(y)-f(x_{i})\rvert<\epsilon$.

We have shown that $\lvert y-x_{i}\lvert<\delta(c_{0})-\dfrac{\bar{\delta}}{2}\Rightarrow \lvert f(y)-f(x_{i})\rvert<\epsilon$, therefore $\delta(c_{0})-\dfrac{\bar{\delta}}{2}\in A(x_{i})$ for large $i$. Hence $E(x_{i})\geq\delta(c_{0})-\dfrac{\bar{\delta}}{2}$ for $i$ greater than a certain $i_{0}$.

Therefore $\inf\{E(x): x\in [a,b]\}=\lim E(x_{i})\geq\delta(c_{0})-\dfrac{\bar{\delta}}{2}>0$, that is $\Delta>0$.

Cosequently $E(x)\geq\Delta>0$ for all $x \in [a,b]$.

So taking a $\delta >0$ such that $\delta<\Delta$ we obtain:

$\lvert y-x\rvert<\delta\Rightarrow\lvert y-x\rvert<\Delta\leq E(x)$. Since $E(x)=\sup A(x)$ there is a $\tilde{\delta}$ arbitrarily close to $E(x)$, hence $\tilde{\delta}\in A(x)$ and $\tilde{\delta}>\delta$, therefore $\lvert y-x\rvert<\delta<\tilde{\delta}$ and hence $\lvert f(x)-f(y)\rvert<\epsilon$.

So there exists a $\delta>0$ such that $\lvert x-y\rvert<\delta \Rightarrow \lvert f(x)-f(y)\rvert<\epsilon$, and since $\epsilon$ was arbitrarily chosen in the first place we have proved uniform continuity!

I will be deeply obliged to anyone who reads this long proof, and I would be indebted if he finds some errors or he has a simpler proof to propose!

Answer 1

You are actually using compactness when you choose a convergent subsequence $x_i \to \overline{x}$. But if you are happy with that kind of argument, then this can be shortened significantly: Assume that $f$ is not uniformly continuous. Then there is an $\varepsilon > 0$ and two sequences $(x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}}$ in $[a,b]$ such that $|x_n-y_n| < \frac{1}{n}$ and $|f(x_n)-f(y_n)| \geq \varepsilon$. By taking subsequences, we may assume that $x_n \to x$ and $y_n \to y$. In fact, $x = y$ because $|x_n-y_n| < \frac{1}{n}$. It follows $$0 = |f(x)-f(y)| = \lim\limits_{n \to \infty} |f(x_n)-f(y_n)| \geq \varepsilon,$$ where we used the continuity of $f$. Contradiction.

Answer 2

Let me provide the standard proof.

Assume that $f: [a,b] \to \mathbb{R}$ is not uniformly continuous. Then: $$\exists \epsilon > 0: \forall \delta >0:\exists x,y \in [a,b]:|x-y|<\delta \land |f(x)-f(y)| \geq \epsilon$$

In other words: $$\forall n \in \mathbb{N}: \exists x_n, y_n \in [a,b]: |x_n-y_n| < \frac{1}{n} \land |f(x_n)-f(y_n)| \geq \epsilon$$ This actually allows us to form two sequences $(x_n)_{n \in \mathbb{N}}$ and $(y_n)_{n \in \mathbb{N}}$. Now, I will leave it to you to try and determine why these two sequences can be assumed to be convergent (Hint: Bolzano-Weierstrass helps you over here). Let the limits of each of these sequences be $x$ and $y$. Then: $$\lim_{n \to \infty} |x-y| \leq \lim_{n \to \infty} |x-x_n| + \lim_{n \to \infty} |x_n-y_n| + \lim_{n \to \infty} |y_n-y| = 0$$ I've purposely skipped steps here so you can fill in some of the details on your own. But this implies that $x = y$. On the other hand, the continuity of $f$ at $x$ entails that: $$\lim_{n \to \infty} |f(x_n)-f(y_n)| = 0 \geq \epsilon$$ and this is a contradiction. Hence, $f$ must be uniformly continuous. $\Box$

Now, it's strange that you classify a compactness argument as being "indirect". Actually, the compactness argument is direct! Indeed, Let $\epsilon > 0$ be given. For each $x \in [a,b]$, we can find a $\delta_x$ such that: $$y \in B(x,\delta_x) \implies f(y) \in B(f(x),\epsilon)$$ In other words, $f(B(x,\delta_x)) \subseteq B(f(x),\epsilon)$. It follows that $\{B(x,\frac{1}{2}\delta_x)\}_{x \in X}$ is an open cover for $[a,b]$. But this means that we can find finitely many points $x_1,\ldots,x_n$ such that $\{B(x_i,\frac{1}{2}\delta_{x_i})\}_{i=1}^{n}$ is a finite subcover of $[a,b]$. Next, choose $\delta = \min \frac{1}{2} \delta_i$ and let $|x-y| < \delta$. Observe that $x \in B(x_i,\frac{1}{2}\delta_{x_i})$. But this means that: $$|y-x_i| \leq |y-x| + |x-x_i| < \frac{1}{2}\delta_{x_i} +\frac{1}{2}\delta_{x_i} = \delta_{x_i}$$

So, $y \in B(x_i,\delta_{x_i})$. But that means that: $$|f(x)-f(y)| \leq |f(x)-f(x_i)| +|f(x_i)-f(y)| < 2\epsilon$$ and that shows what we wanted. This is direct! $\Box$

I actually wanted to make two remarks on the first proof:

1. In that proof, it may not look like it but you are actually making use of a compactness principle. Indeed, the Bolzano-Weierstrass Theorem is a compactness principle on the real line. It's just that it has more to do with sequential compactness.

2. In the moment where I formed those two sequences, I actually did make use of the axiom of choice. The proof there actually isn't constructive; I haven't given a direct, explicit way to find those two sequences. The compactness argument (via open covers) circumvents this. Having said that, my favorite argument is actually the sequential one because it is very simple.