Proof Verification Regarding Uniform Continuity Assume that $g$ is defined on an open interval $(a, c)$ and it is known to be uniformly continuous on $(a, b]$ and $[b, c)$, where $a < b < c$. Prove that g is uniformly continuous on $(a, c)$.
My attempt at proof:
By hypothesis, we have:
$$\forall \epsilon>0, \exists \delta_1>0, \forall x,y \in (a,b]: |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon. $$ and $$\forall \epsilon>0, \exists \delta_2>0, \forall x,y \in [b,c): |x-y|<\delta_2 \implies |f(x)-f(y)|<\epsilon. $$
If we pick $\delta = \min \{\delta_1,\delta_2 \}$, then we have $$\forall \epsilon>0, \exists \delta>0: |x-y|<\delta \implies |f(x)-f(y)|<\epsilon, $$ whenever $x,y$ is in $(a,b]$ or $[b,c)$, or equivalently, whenever $x,y$ is in $(a,c)$.
Is this correct?
 A: I think the idea of your proof is right.  Basically, when proving continuity (or uniform continuity), given $\epsilon > 0$, if you can find a $\delta > 0$, then any smaller $\delta '$ will also work.  I just think you need to be a bit more careful about the details.
So we know given $\epsilon > 0$, we can find $\delta_{1} > 0$ such that for all $x, y \in (a,b]$ (even if one of them equals $b$), $|x - y| < \delta_{1} \implies |f(x)- f(y)| < \frac{\epsilon}{2}$, right?
Also, for the same $\epsilon > 0$, we can find $\delta_{2} > 0$ such that for all $w, z \in [b,c)$ (in particular, also if one of them equals $b$), $|x - y| < \delta_{2} \implies |f(x) - f(y)| < \frac{\epsilon}{2}$.
Now, if both of our inputs are in $(a,b]$, then letting $\delta = \min \{\delta_{1}, \delta_{2} \}$ is OK because any smaller $\delta$ still works.  Similarly, if they are both in $[b,c)$, it still works.  We only have to worry about if $x \in (a,b]$ and $y \in [b,c)$ (since we want $x,y \in (a,c)$ to be arbitrary, so we have to consider all cases).  So let's assume this case since the other cases are taken care of.
So, let's assume $x \in (a,b]$ and $y \in [b,c)$.  Then it should be clear that the distance between $x$ and $b$ is smaller than the distance between $x$ and $y$, right?  i.e., $|x - b| \leq |x - y|$.  Similarly, the distance between $y$ and $b$ is smaller than the distance between $x$ and $y$, i.e., $|y - b| \leq |x - y|$.  Then if $|x - y| < \delta$, we have $|x - b|$, $|y - b| < \delta$, right?  Then the first one is smaller than $\delta_{1}$, and the second one is smaller than $\delta_{2}$, right?  So $|f(x) - f(b)| < \frac{\epsilon}{2}$ and $|f(y) - f(b)| < \frac{\epsilon}{2}$.
So, if $|x - y| < \delta$, the above inequalities hold.  Finally,  we want to use the trick of adding $0$ in the form of $-f(b) + f(b)$ to show that $ |f(x) - f(y)| < \epsilon$ if $|x - y| < \delta$.
So, given $\epsilon > 0$, let $\delta = \min \{ \delta_{1}, \delta_{2} \}$.  Then $|x - y| < \delta$ implies $ |f(x) - f(y)| = |f(x) -  f(b) + f(b) - f(y)| \leq |f(x) - f(b)| + |f(b) - f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$, and we are done.
