Suppose that the function $f:[0,1)\rightarrow \mathbb{R}$ is uniformly continuous on $[0,1).$ Prove that the function $f$ is bounded. (i.e. that the range$(f)$ is a bounded set)
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How does the endpoint change the proof? – user296855 Dec 08 '15 at 04:37
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Just so you're aware, you accept answers by clicking the check mark just below the downvote arrow. – AnotherPerson Dec 08 '15 at 08:49
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If a function $f:[0,1)\to \mathbb{R}$ is uniformly continuous, then for any $\epsilon>0$, there is a $\delta$ such that if $x, y\in [0,1)$ are $\delta$ apart, they will have images in $\mathbb{R}$ that will be within $\epsilon$ of each other i.e. $|f(x)-f(y)|<\epsilon$.
Now since the endpoints in our domain are a finite distance apart (namely everything in the domain has distance less than 1 apart), we have that the maximum height difference between $f(x)$ and $f(y)$ must be the $\epsilon$ corresponding to $\delta=1$. Given that the distance between any two values in the range must be less than this $\epsilon$, we have that our range is bounded, and hence so is $f$.
AnotherPerson
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Since $f$ is uniformly continuous, $\exists \delta$ such that $|f(x)-f(y)|<\delta |x-y|$ for all $x,y\in [0,1)$. Since $|x-y|<1$ for all $x,y\in [0,1)$, $|f(x)-f(y)|<\delta\cdot(1)=\delta$. Thus the range is bounded.
Jürgen Sukumaran
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Uniform continuity of $f$ on $[0,1)$ means I can extend $f$ continuously to $[0,1]$. Then, this continuous extension will have compact image of $[0,1]$ since it is continuous and $[0,1]$ is compact. So this image, $\tilde f([0,1])$ will contain it's sup and inf. Then this sup and inf will also be valid for $f$ on $[0,1)$ and so the image of $f$ on $[0,1)$ will be bounded by these values. – Jürgen Sukumaran Dec 08 '15 at 04:38