Assume $g$ is defined on $(a,c)$ and uniformly continuous on $(a,b]$ and $[b,c)$ with $a<b<c$

Question

Problem:
Assume $g$ is defined on $(a,c)$ and uniformly continuous on $(a,b]$ and $[b,c)$ with $a<b<c$.

Show that $g$ is uniformly continuous on $(a,c)$

I think only case we have to consider is when $x$ in $(a,b]$ and $y$ in $[b,c)$ since if $x,y \in (a,b]$ (or the other) then we are done.

How can I prove this using epsilon and delta ?

Answer 1

Hint: If you find a $\delta$ that works with tolerance $\epsilon/2$ for both $(a,b]$ and $[b,c)$, then...

Answer 2

WLOG assume $x<y$. Since $f$ is uniformly continuous on $(a,b]$ and $[b,c)$, given $\epsilon>0$, there is a $\delta>0$, for any $x<y\in (a,b]$ and $|x-y|<\delta$, as well as any $x<y\in [b,c)$ and $|x-y|<\delta$, there is $|f(x)-f(y)|<\epsilon$.

If $x\in(a,b]$ and $y\in[b,c)$ that $|x-y|<\delta$, then $|x-b|<\delta$ and $|y-b|<\delta$. So $|f(x)-f(b)|<\epsilon$ and $|f(y)-f(b)|<\epsilon$. Thus $$ |f(x)-f(y)|=|f(x)-f(b)+f(b)-f(y)|\leqslant |f(x)-f(b)|+|f(b)-f(y)|<2\epsilon $$ i.e. $f$ is uniformly continuous on $(a,c)$.