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Prove: If a function $f: (a,b) \to\mathbb{R}$ is uniformly continuous, then $f(a+)$ and $f(b-)$ are both finite.

I think that the best way for this to be proved is by contradiction...

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    I think the best way is Cauchy convergence principle.2012-04-12

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let $x_n \in (a,b)$, $x_n \to a$. For $\epsilon > 0$ there is an $\delta > 0$ such that $|f(x) - f(y)| < \epsilon$, given $|x-y| < \delta$. Choose $N$ with $|x_n - x_m| < \delta$ for $n, m \ge N$. Then $|f(x_n) - f(x_m)| \le \epsilon$ for these $n, m$. As $\epsilon$ was arbitrary, $(f(x_n))$ is Cauchy, thus convergent. Therefore $f(a+)$ exists.

Hope this helps,