# Suppose $-\infty \le a < c < b \le ∞$ and $f:(a,b)\to \mathbb{R}$ is continuous on $(a,b)$.

Suppose $-\infty \le a < c < b \le ∞$ and $f:(a,b)\to \mathbb{R}$ is continuous on $(a,b)$.

(a) If $f$ is uniformly continuous on both $(a,c)$ and $(c,b)$, prove that $f$ is uniformly continuous on $(a,b)$.

(b) Give an example to show that the conclusion in part (a) may be false if $f$ is not continuous on $(a,b)$.

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The problem with the way this question is phrased is that it looks as if "doit" is passing on to us a question written by someone other than "doit", so it's not clear whether the poster even understands the question. –  Michael Hardy May 16 '12 at 22:01
@MichaelHardy I actually did not look at the questions posted by "doit" earlier. Now looking at the previous questions posted by "doit", Arturo has already left comment on a previous question of the user here (math.stackexchange.com/questions/144461/…) –  user17762 May 16 '12 at 22:04
The imperative seems to be connected to the username... –  Michael Greinecker May 16 '12 at 22:06
Let $\epsilon\gt 0$. You know that there exists $\delta_0$ such that any two points that are $\delta_0$ close and both are in $(a,c)$ will have images that are $\epsilon$-close. And you know that there is a $\delta_1$ such that any two points that are $\delta_1$ close and both are in $(c,d)$ will have images that are $\epsilon$-close. Taking the minimum of $\delta_1$ and $\delta_2$ will guarantee any two points that are in the same subinterval and are within $\delta$ of each other will have images that are $\epsilon$-close. The problem lies with the situation in which one point is in $(a,c)$ and the other point is in $(c,b)$.
Now, we know the function is continuous at $c$, so we also know there is a $\delta_2$ such that if a point is within $\delta_2$ of $c$, then its image is within $\epsilon$ of the image of $c$.