$f$ is uniformly continuous on $[a,b]$ and $[b-1,c]$ $\Rightarrow$ uniformly continuous on $[a,c]$

Question

$f$ is uniformly continuous on $[a,b]$ and $[b-1,c]$ $\Rightarrow$ uniformly continuous on $[a,c]$

My thoughts:

Without loss of generality we only need to show that it's uniform if $x_1 \in [a,b]$ and $x_2 \in [b-1,c]$ (since we already know it's uniform if $x_1,x_2 \in $ the same set).

Using the definition of uniform continuity we know:

$|a-b|<\delta _1 \Rightarrow |f(a)-f(b)|<\epsilon /2$ for $[a,b]$

$|c-(b-1)|<\delta _2 \Rightarrow |f(c)-f(b-1)|<\epsilon /2$ for $[b-1,c]$

Combining we get:

$$|x_1-x_2|<|a-b|+|c-(b-1)|<\delta _1 +\delta _2 \Rightarrow |f(a)-f(b)|+|f(c)-f(b-1)|<\epsilon$$

Is this correct?

Answer 1

Let $I_1=[a,b], I_2=[b-1,c]$. Let $x\in I_1$, $z\in I_2$ and $y \in I_1\cap I_2$.

Given $\varepsilon>0$, we know there exists $\delta_1,\delta_2$ such that for every $x,y,z$ (in the respective domains) we have,

\begin{align} |x-y|&<\delta_1 \rightarrow |f(x)-f(y)|<\frac \varepsilon 2\\ |y-z|&<\delta_2 \rightarrow |f(y)-f(z)|<\frac \varepsilon 2 \end{align}

Now,

$$|x-z|\leq |x-y|+|y-z|<\delta_1+\delta_2 \rightarrow|f(x)-f(y)|+|f(y)-f(z)|<\varepsilon$$

But also, $|f(x)-f(z)|\leq |f(x)-f(y)|+|f(y)-f(z)|$ by triangle inequality.