# 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.

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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$here) – Beni Bogosel Mar 17 '13 at 20:44 – 1015 Mar 17 '13 at 20:55 This theorem illustrates the importance of distinguishing very large and infinite. Your example gives a direct illustration of this, when coupled with the theorem. – Kieran Cooney Oct 26 '13 at 18:51 ## 2 Answers I can provide you with a proof. We use the lemma that$[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$|b-x|<\delta$, 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, take any pair$x,y\in[a,c]$. If$x,y\in[a,b]$or$x,y\in[b,c]$, we're done. If$x<b<y$or$y<b<x$. 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 <b$. Since$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$\delta =\min{d_{x_i}}$, and let$x,y\in [a,b]$with$d(x,y)<\delta$. Since$\mathcal O$is a cover, for some$x_i$we have that$|x-x_i|<\delta_{x_i}$. Then, we'll have $$|y-x_i|\leq |y-x|+|x-x_k|<\delta+\delta_i\leq 2\delta_i$$ It follows that $$|f(x_i)-f(x)|<\epsilon/2$$ $$|f(y_i)-f(x)|<\epsilon/2$$ 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, one can give the 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.

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