# A continuous function $f$ from a closed bounded interval $[a, b]$ into $\mathbb{R}$ is uniformly continuous

I am reading (as a supplement) the book Basic Real Analysis, by Anthony Knapp. Before I proceed into reading a proof, I want to be sure that the result seems obvious. Yet, I am having trouble seeing through this one. It annoys me too much in order to disregard it:

Theorem. A continuous function $f$ from a closed bounded interval $[a, b]$ into $\mathbb{R}$ is uniformly continuous.

What gives? Why can't we provide the counterexample $f(x)=x^2$ and $[a,b] \subset [1,+\infty)$, for $b < +\infty$ sufficiently large and show that the theorem is incorrect?

Doesn't it seem to be an insufficient statement?

I'm having trouble picturing it, is all.

Hints are fine.

• Because even if you take $b$ as large as you want, it is still finite, and you still have $|f(x)-f(y)| \leq C|x-y|$ for a big constant $C$. (I'm talking about the function f(x)=x^2$here) – Beni Bogosel Mar 17 '13 at 20:44 • – Julien Mar 17 '13 at 20:55 • This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem. – Kieran Cooney Oct 26 '13 at 18:51 ## 2 Answers I can provide you with a proof. We use the lemma that $$[a,b]$$ is compact. The general statement is that if $$f:X\to Y$$ is continuous and $$X$$ is a compact metric space, then i$$f$$ is uniformly continuous. This is usually known as the Heine Cantor theorem, while the fact that $$[a,b]$$is compact might be found as Borel's Lemma, if memory serves. So THEOREM (Spivak) Let $$f:[a,b]\to \Bbb R$$ be continuous. Then it is uniformly continuous. We first prove the LEMMA Let $$f$$ be a continuous function defined on $$[a,c]$$. If, given $$\epsilon >0$$, there exists $$\delta_1>0$$ such that, for each pair $$x,y\in[a,b]\text{ ; } |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon$$ and $$\delta_2>0$$ such that for each $$x,y\in[b,c]\text{ ; } |x-y|<\delta_2 \implies |f(x)-f(y)|<\epsilon$$ Then there exists $$\delta$$ such that for each $$x,y\in[a,c]\text{ ; } |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$$ P Since $$f$$ is continuous at $$x=b$$, there exists a $$\delta_3$$ such that for every $$x$$ with $$|b-x|<\delta_3$$, we have $$|f(b)-f(x)|<\frac{\epsilon}2$$. Thus, whenever $$|x-b|<\delta_3$$ and $$|y-b|<\delta_3$$ we will certainly have $$|f(x)-f(y)|<\epsilon$$ We take $$\delta=\min\{\delta_1,\delta_2,\delta_3\}$$. Then $$\delta$$ works: indeed, consider any pair $$x,y\in[a,c]$$. If $$x,y\in[a,b]$$ or $$x,y\in[b,c]$$, we're done. If $$x or $$y. In any case, since $$|x-y|<\delta$$, we must have $$|x-b|,|y-b|<\delta$$, so that $$|f(x)-f(y)|<\epsilon$$, as claimed. PROOF1 Fix $$\epsilon >0$$. Let's agree to call $$f$$ $$\epsilon$$-good on an interval $$[a,b]$$ if for this $$\epsilon$$ there exists a $$\delta$$ such that for any $$x,y\in[a,b]$$, $$|x-y|<\delta\implies |f(x)-f(y)|<\epsilon$$. We thus want to prove that $$f$$ is $$\epsilon$$-good on $$[a,b]$$ for any $$\epsilon >0$$. Let $$\epsilon >0$$ be given, and consider the set $$A(\epsilon)=\{x\in[a,b]:f \text{ is } \epsilon \text{-good on}: [a,x]\}$$ Then $$A\neq \varnothing$$ for $$a\in A(\epsilon)$$, and $$A(\epsilon)$$ is bounded above by $$b$$. Thus $$\sup A=\alpha$$ exists. Suppose that $$\alpha . Since $$f$$ is continuous at $$\alpha$$ there exists a $$\delta'$$ such that $$|y-\alpha|<\delta'$$ implies $$|f(y)-f(\alpha)|<\epsilon/2$$. Thus, if $$|y-\alpha|,|x-\alpha|<\delta'$$, we'll have $$|f(y)-f(x)|<\epsilon$$. Thus $$f$$ is $$\epsilon$$-good on $$[\alpha-\delta,\alpha+\delta]$$. Since $$\alpha=\sup A(\epsilon)$$, it is clear $$f$$ is $$\epsilon$$-good on $$[a,\alpha+\delta]$$, which is absurd. Thus $$\alpha\geq b$$, which means $$\alpha =b$$. It suffices to show that $$b$$ is also an element of $$A(\epsilon)$$. But since $$f$$ is continuous on $$b$$, there exists a $$\delta_0$$ such that $$|b-y|<\delta_0$$ implies $$|f(b)-f(y)|<\epsilon/2$$. Thus, $$f$$ is $$\epsilon$$-good on $$[b-\delta_0,b]$$. The lemma implies $$f$$ is $$\epsilon$$-good on $$[a,b]$$. Since $$\epsilon$$ was arbitrary, the result follows. $$\blacktriangle$$ PROOF2 Let $$\epsilon >0$$ be given. Assign, to each $$x\in [a,b]$$ a $$\delta_x>0$$ such that for each $$y\in(x-2\delta_x,x+2\delta_x)$$, we have $$|f(x)-f(y)|< \epsilon/2$$, to obtain a open cover of $$[a,b]$$, namely the set $$\mathcal O$$ of intervals $$(x-\delta_x,x+\delta_x)$$. This is possible since $$f$$ is continuous at each $$x$$. Since $$[a,b]$$ is compact, there is a finite number of $$x_i\in [a,b]$$ such that $$\bigcup_{i=1}^n (x_i-\delta_{x_i},x_i+\delta_{x_i})\supset [a,b]$$ Choose now $$\delta =\min{\delta_{x_i}}$$, and let $$x,y\in [a,b]$$ with $$|y-x|<\delta$$. Since $$\mathcal O$$ is a cover, for some $$x_i$$ we have that $$|x-x_i|<\delta_{x_i}$$. Then, we'll have $$|y-x_i|\leq |y-x|+|x-x_i|<\delta+\delta_i\leq 2\delta_i$$ It follows that $$|f(x_i)-f(x)|<\epsilon/2$$ $$|f(y)-f(x)|<\epsilon/2$$ which means by the triangle inequality that $$|f(x)-f(y)|<\epsilon$$ Then for any $$x,y\in[a,b]$$, $$|x-y|<\delta$$ will imply $$|f(x)-f(y)|<\epsilon$$; and $$f$$ is uniformly continuous. $$\blacktriangle$$ There is yet another way of proving this. LEMMA (Mendelson) Let $$X$$ be a metric space such that every infinite subset of $$X$$ has an accumulation point in $$X$$. Then for each covering $$\mathcal O=\{O_\beta\}_{\beta\in I}$$ there exists a positive $$\epsilon$$ such that each ball $$B(x;\epsilon)$$ is contained in an element $$O_\beta$$ of this covering. PROOF If it wasn't the case, we'd obtain for each $$n$$ an point $$x_n$$ and an open ball $$B(x_n;1/n)\not\subseteq O_\beta$$ for each $$\beta \in I$$. Let $$A=\{x_1,\dots\}$$. If $$A$$ is finite, $$x_n=x$$ infinitely often for some $$x\in X$$. Since $$\mathcal O$$ is a cover, $$x\in O_\alpha$$ for some $$\alpha$$. Since the cover is open, there is a $$\delta >0$$ for which $$B(x;\delta)\subseteq O_\alpha$$. We can take $$n$$ such that $$1/n<\delta$$ and $$x_n=x$$, in whichcase we get a contradiction $$B\left(x;\frac 1n \right)\subseteq B\left(x;\delta\right)\subseteq O_\alpha$$ If $$A$$ is infinite, there is an accumulation point $$x\in X$$. Thus $$x\in O_\beta$$ for some index, and there are infinitely many points of $$A$$ in $$B(x:\delta /2)\subseteq O_\beta$$. We can take $$n$$ such that $$1/n<\delta /2$$ and we'd have $$B(x_n;1/n)\subseteq B(x;\delta)\subseteq O_\beta$$, a contradiction. After having proven that for metric spaces, the existence of accumulation points for infinite subsets is equivalent to compactness. PROOF3 Let $$f:X\to Y$$ be a continuous function from a compactum $$X$$ to a metric space $$Y$$. Then $$f$$ is uniformly continuous. PROOF Given $$\epsilon >0$$, for each $$x\in X$$ there is a $$\delta_x>0$$ such that if $$y\in B(x:\delta_x)$$, $$f(y)\in B\left(f(x);\epsilon /2\right)$$. These balls are an open cover for $$X$$, thus there exists such a number $$\delta_L$$ as in the previous lemma (usually called a Lebesgue number). Choose $$\delta$$ to be positive yet smaller than $$\delta_L$$. If $$z,z'\in X$$ and $$d(z,z')<\delta$$ (so that $$z,z'$$ are in a ball of radius less than $$\delta$$), we have $$z,z'\in B(x,\delta_x)$$ for some $$x\in X$$. In that case $$f(z),f(z')\in B(f(x),\epsilon/2)$$ so $$d'(f(z),f(z'))<\epsilon$$ by the triangle inequality. $$\blacktriangle$$. • I think you need to elaborate more on the last step.$\delta_x$and$\delta_y$could be smaller than the chosen$\delta$. You need to find the$x_i$for which$x \in (x_i - \delta_{x_i}, x_i + \delta_{x_i})$, and make sure that$y$is close enough to$x$so that it also belongs to this interval. – Ayman Hourieh Mar 18 '13 at 9:33 • @Ayman I see what you mean. – Pedro Tamaroff Mar 18 '13 at 13:11 • Thanks. The elaboration needed for your original proof isn't big. It's something along the lines of my previous comment. But it is necessary; otherwise it's unclear why$\left|f(x) - f(y)\right| < \epsilon\$. – Ayman Hourieh Mar 18 '13 at 17:40
• @ayman After some reading I have found something that fixes my proof try. I will try and digest it, and if possible add it tomorrow. – Pedro Tamaroff Mar 26 '13 at 2:29
• The fix I had in mind was similar to proof 1 at proofwiki.org/wiki/Heine-Cantor_Theorem#Proof_1. – Ayman Hourieh Mar 26 '13 at 15:43

In this case, intuitively, no matter how large the interval is, as long as it is closed and bounded, the function is still controlled. This is a special case of a more general result: a continuous function on a compact metric space is uniformly continuous, and we know that a subset of Euclidean space is compact if and only if it is closed and bounded.