If I have function $f$ which is uniformly continuous at interval $[a,b)$ and at interval $[b, c)$ as well, can I conclude that $f$ is uniformly continuous at $[a, c)$?
I am pretty sure that this argue is true, but I can't find a proper proof. What I though about was taking the minimum value of $\delta_1$ and $\delta_2$ (which are the working deltas for $[a,b)$ and $[b, c)$) and say that for that delta the function if uniformly continuous at the whole interval. However, I do not manage to prove it.
Thanks
This is not necessarily true. Consider $$f(x)=\begin{cases}0 & x< 1\\ 1 & x\geq 1\\\end{cases}.$$ If $a=0$, $b=1$, and $c=2$, we can conclude that $f$ is uniformly continuous of $[a,b)$ and $[b,c)$, but not on $[a,c)$. However, the idea you have of taking the minimum of two $\delta$ values is a good one. If you insist that $f$ also be continuous at $b$, the method you are thinking of will work. It goes like this:
Suppose that $f$ is uniformly continuous on $[a,b)$ and $[b,c)$ and is continuous at $b$. Fix $\varepsilon>0$. Then there exists $\delta_1$ and $\delta_2$ which satisfy the definition of uniform continuity in $[a,b)$ and $[b,c)$ respectively. There also exists a $\delta_3$ such that if $|x-b|<\delta_3$, then $|f(x)-f(b)|<\varepsilon/2$. Let $\delta=\min\{\delta_1,\delta_2,\delta_3\}.$ Choose $x,y\in [a,c)$ such that $|x-y|<\delta$. Without loss of generality, assume $x<y$. Clearly, if $x,y\in [a,b)$, then $|x-y|<\delta\leq\delta_1$, so $|f(x)-f(y)|<\varepsilon$. Similarly if $x,y\in [b,c)$. If $a\leq x<b<y<c$, Then $|x-b|<|x-y|<\delta\leq\delta_3$, so $|f(x)-f(b)|<\varepsilon/2$. Similarly, $|f(b)-f(y)|<\varepsilon/2$. Thus, $$|f(x)-f(y)|\leq |f(x)-f(b)|+|f(b)-f(y)|<\varepsilon/2+\varepsilon/2<\varepsilon.$$ So we see that in all cases, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$, so $f$ is uniformly continuous on $[a,c)$.