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...
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...
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,