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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.

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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 $x_0$.

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  • $\begingroup$ So what deltas would we pick? $\endgroup$ – user259137 Aug 6 '15 at 19:22
  • $\begingroup$ I want to leave that for you to figure out yourself. I was just trying to give you the big picture here. $\endgroup$ – Alex G. Aug 6 '15 at 20:42

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