# Thread: Piecewise continuous

1. ## Piecewise continuous

Let $\displaystyle D=[0,1] \cup (2,3]$ and define $\displaystyle f: D \rightarrow \mathbb{R}$ by $\displaystyle f(x) = \left\{\begin{array}{cc}x,&\mbox{ if } 1 \leq x\leq 0\\x-1, & \mbox{ if } 2<x\leq 3\end{array}\right.$

Prove that f is continuous.

The problem is the disconnected domain, if I pick any point $\displaystyle x_0 \in [0,1]$, would I pick $\displaystyle x \in [0,1]$ with $\displaystyle |x-x_0| < \delta$, then process with the proof, then do the same for (2,3]?

2. Hello,
Originally Posted by tttcomrader
Let $\displaystyle D=[0,1] \cup (2,3]$ and define $\displaystyle f: D \rightarrow \mathbb{R}$ by $\displaystyle f(x) = \left\{\begin{array}{cc}x,&\mbox{ if } 1 \leq x\leq 0\\x-1, & \mbox{ if } 2<x\leq 3\end{array}\right.$

Prove that f is continuous.

The problem is the disconnected domain, if I pick any point $\displaystyle x_0 \in [0,1]$, would I pick $\displaystyle x \in [0,1]$ with $\displaystyle |x-x_0| < \delta$, then process with the proof, then do the same for (2,3]?
Yes, and you'll prove that f is continuous separately on $\displaystyle [0,1]$ and $\displaystyle (2,3]$