I need some help on a problem. In my real analysis class, we have just been introduced to the idea of uniform continuity. While I know how to show that $\mathit{f(x) = x}$ is continuous, I need some help showing that $\mathit{f(x) = x }$is uniformly continuous. Would someone mind taking a look at what I have done so far? Thanks in advance.
So far this is what I have:
Let $\mathit{f:S\rightarrow \mathbb{R}}$, $\mathit{S}\subseteq\mathbb{R}$, be defined by $\mathit{f(x)=x}$.
Then, $\forall \epsilon > 0, \exists \delta > 0$ such that $\forall$ $x_{0}$ $\in \mathit{S}$ and $\forall$ x $\in \mathit{S}$ such that |$\mathit{x} - \mathit{x_{0}}$| $\Rightarrow$ |$\mathit{f(x)} - \mathit{f(x_{0})}$|.
Thus, we can conclude that $\mathit{f(x) = x}$ is uniformly continuous.
I went back and worked through this again with the comments below in mind. How does this look?
Let $\mathit{f:S\rightarrow \mathbb{R}}$, $\mathit{S}\subseteq\mathbb{R}$, be defined by $\mathit{f(x)=x}$.
We know that for $\mathit{f}$ to be continuous, then $\forall \epsilon > 0$, $\exists \delta > 0$ such that if $x_{0}$ and $\mathit{x}$ $\in \mathit{S}$ and |$\mathit{x} - \mathit{x_{0}}$| < $\epsilon$, |$\mathit{f(x)} - \mathit{f(x_{0})}$| < $\delta$ must hold.
Let $\epsilon$ = $\delta$.
Then, whenever |$\mathit{x} - \mathit{x_{0}}$| < $\delta$, then |$\mathit{f(x)} - \mathit{f(x_{0})}$| < $\epsilon$ = $\delta$.
Thus, $\mathit{f(x)=x}$ is uniformly continuous on $\mathbb{R}$.
Your proof seems like you have the right idea, but just needs a little better wording.
Indeed, $\forall \varepsilon > 0$, there does exist a $\delta$. In fact, you should say that we choose $\delta = \varepsilon$. This choice of delta thus does not depend on the values $x,x_0\in S$ that you choose.
Then we see that whenever $|x-x_0|<\varepsilon$, we have $|f(x)-f(x_0)|=|x-x_0|<\varepsilon=\delta$.