# Using $\epsilon$-$\delta$ argument to show continuity

Show using the $\epsilon$-$\delta$ definition of continuity that $f(x)=\begin{cases} 11&\text{if}~0\leq x\leq 1\\x&\text{if}~ 1<x\leq 2\end{cases}$ is continuous on $[0,1)\cup (1,2]$

How do we construct this proof, I am confused by the half closed sets as well as $f(x)$ being $11$, a constant.

You should think of the half closed sets in this way: $[0, 1)\cup (1, 2]$ is the set of all numbers between $0$ and $2$, inclusive, except for $1$. So, what you need to do is show that the given function is continuous everywhere except for $1$.
Here's a hint: Take $x_0$ to be the point at which we're trying to show $f(x)$ is continuous. If $x_0 < 1$, then for all points sufficiently close to $x_0$, $f(x)$ is just $11$. Thus $|f(x) - f(x_0)| = |11-11| = 0$ for $x$ sufficiently close to $x_0$.
On the other hand, if $x_0 > 1$, then for points $x$ sufficiently close to $x_0$, $f(x) = x$. Thus $|f(x) - f(x_0)| = |x-x_0|$, which is small if $x$ is close to $x_0$.