In some textbooks, sequences with a weaker form of Axiom of choice (sometimes) are excessively used to prove many theorems like "Boundness theorem", "Extreme Value theorem" and "Root location theorem", but they can be proven without using sequences as I found in this article in Wikipedia:

However, it is not the same with "Uniform Continuity theorem" which states that "any continuous function on a closed interval is a uniform continuous on the same interval". I attempt define the following set given that $f:[a,b] \to \mathbb{R}$ is continuous on $[a,b]$

$$ S = \left\{ x \in [a,b]\; \middle| \; f\; \text{is uniformly continuous on }[a,x]\right\} $$

Here, I intent to use the same style that in the same article. It is easy to show that $a \in S$, but I cannot prove that $a + \delta$ (where $\delta$ is a positive number) is in S (or $f$ is uniformly continuous on $[a,a+\delta]$ from the fact that $f$ is continuous on $a$ or anything else, nor prove that $b = \sup(S)$

I googled "uniform continuity theorem" and the only proof I found is the sequential one which I want to avoid.

So, is it even possible to prove it without sequences?


2 Answers 2


You can, in fact, prove uniform continuity in this style… with a slight modification.

(The flaw in Cactus’s counter-argument – though their proof is the “better” one – is that even for a function $\mathbb{R} \rightarrow \mathbb{R}$, if $S=\mathbb{R}$, it only means that $f$ is uniformly continuous on every closed bounded interval.)

The issue with your method is that you’re more or less reduced to proving that $f$ is uniformly continuous on $[a,a+\delta]$ – which is hopeless, because it’s already the statement that you want to prove!

The trick is to extract $\epsilon$ from $S$.

Fix $\epsilon >0$, and let $S$ be the set of $x \in [a,b]$ such that there exists $\eta>0$ such that for all $u,v \in [a,x]$ with $|u-v| \leq \eta$, $|f(u)-f(v)| \leq \epsilon$.

Now, extending $S$ from $[a,c]$ to $[a,c+\delta]$ of from $[a,c)$ to $[a,c]$ is just the continuity of $f$ at $c$.

  • $\begingroup$ Thanks a lot, but why $|u - v| \le \eta$ instead of $|u - v| < \eta$ (the same with $|f(u) - f(v)| \le \epsilon$)? $\endgroup$ Jun 9 at 7:05
  • 1
    $\begingroup$ Usually those are up to preference, as you could notch up/down the bounds if needed. $\endgroup$
    – Abastro
    Jun 9 at 10:08

You can't do it like that. Indeed, if such an argument worked, you could use the same reasoning to prove that any function $\mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous on $\mathbb{R}$ (which is wrong, take $x \mapsto x^2$).

If you want to avoid sequences, you can use the Borel-Lebesgue property on compact sets : for compact set $K$ (think of a closed bounded interval if you don't know what it a compact set), if $(U_i)_{i \in I}$ is a family of open subsets of $K$ (think $U_i$ of the form $]a_i,b_i[ \cap K$ in the interval case) such that $K = \bigcup_{i \in I} U_i$, then there exists a finite subset $J$ of $I$ such that $K = \bigcup_{i \in J} U_i$.

Now, let $\varepsilon > 0$ and $x \in K$. $f$ is continuous at $x$ so there exists a $\delta_x > 0$ such that $d(f(x),f(y)) \leqslant \varepsilon/2$ when $y \in K$ and $d(x,y) \leqslant \delta_x$ (here, $d$ denotes the distances. In the real case, $d(x,y) = |x - y|$). Let for all $x$, $U_x = ]x - \delta_x/2,x + \delta_x/2[ \cap K$. It is clear that $K = \bigcup_{x \in K} U_x$, hence there is a finite set $J \subset K$ such that $K = \bigcup_{x \in J} U_x$.

As $J$ is finite, we can define $\delta = \min_{x \in J}\{\delta_x/2\} > 0$ and if $(x,y) \in K^2$ are such that $d(x,y) \leqslant \delta$, consider some $z \in J$ such that $x \in U_z$. $d(y,z) \leqslant d(x,y) + d(x,z) \leqslant \delta + \delta_z/2 \leqslant \delta_z$ and $d(x,z) \leqslant \delta_z/2 \leqslant \delta_z$.

We deduce that $d(f(x),f(z)) \leqslant \varepsilon/2$ and $d(f(y),f(z)) \leqslant \varepsilon/2$ hence $d(f(x),f(y)) \leqslant \varepsilon$. It proves the uniform continuity of $f$ on $K$.


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