Prove that if $f$ is uniformly continuous on $[a,b]$ and $f$ is uniformly continuous on $[b,c]$, then $f$ is uniformly continuous on $[a,c]$. Prove that if $f$ is uniformly continuous on $[a,b]$ and $f$ is uniformly continuous on $[b,c]$, then $f$ is uniformly continuous on $[a,c]$.
My attempt:
Let $\epsilon >0$ be given. Since $f$ is uniformly continuous on $[a,b]$ there $\exists \delta_1 >0$ such that if $x,y \in [a,b]$ and $|x-y|<\delta_1$, then $|f(x)-f(y)|<\epsilon$.
Since  $f$ is uniformly continuous on $[b,c]$ there $\exists \delta_2 >0$ such that if $x,y \in b,c]$ and $|x-y|<\delta_2$, then $|f(x)-f(y)|<\epsilon$.
Now to show $f$ is continuous on $[a,c]$ how would i show this. Do i sort of add the two above relations? 
 A: Uniform continuity is not a pointwise, relative but an absolute attribute meaning "$f$ is uniformly continuous on $[a,b]$" may be put as "$f$ is uniformly continuous as function on $[a,b]$".
Let $\epsilon>0$ be arbitrary.
Since $f$ is uniformly continuous as function on $[a,b]$ there is a $\delta_1>0$ so that for all $x,y \in [a,b]$: $$|y-x|<\delta_1\implies |f(y)-f(x)|<\epsilon /2.$$ 
Since $f$ is uniformly continuous as function on $[b,c]$ there is a $\delta_2>0$ so that for all $x,y \in [b,c]$: $$|y-x|<\delta_2\implies |f(y)-f(x)|<\epsilon /2.$$
Set $\delta := \min\{\delta_1,\delta_2\}.$
Consider $x,y\in[a,c]$:
If $x,y$ are both in either $[a,b]$ or $[b,c]$ we're done due to $\delta \leq \delta_1$ resp. $\delta \leq \delta_2$.
So without loss of generality it suffices to consider $x\in [a,b]$ and $y\in[b,c]$ for the rest. We then have for $|y-x|<\delta$:
$$|f(y)-f(x)|\leq|f(y)-f(b)|+|f(b)-f(x)| < \epsilon/2+\epsilon/2 = \epsilon.$$
A: Choose the minimum of the deltas from either interval corresponding to epsilon. This delta works.
A: Since f(x) is continuous on [a,b] and [b,c], we have
$$lim_{x\to b^+} f(x)= lim_{x\to b^-} f(x) = f(b)$$
Which implies that f(x) is continuous at x=b. Hence, f(x) is continuous on [a,c], therefore it is uniformly continuous on [a,c].
