Using only the definition of uniform continuity, prove that the following functions are uniform continuous:

Question

Using only the definition of uniform continuity, prove that the following function is uniformly continuous:
$g:[1,2] \to \mathbb R $ be defined by $g(x)=\sqrt x$ for all $x \in [1,2] $ Below is what I have so far. Am I anywhere close?

The function f is uniformly continuous if for each $ε>0$ there is some $δ>0$ such that $x,y∈A$ and $|x-y|<δ$ implies $|f(x)-f(y)|>ε$.
Let $g:[1,2] \to \mathbb R$ be defined by $g(x)=\sqrt x$ for all $x \in [1,2]$, where $x,y \in \mathbb R$.
If we are looking for $ε>0$ and we are given the domain, then we can substitute. $g(x)=\sqrt 2$ and $g(y)=\sqrt 1$. Then, by definition, $|x-y|<δ=|2-1|<δ$ implies $|g(x)-g(y)|=|\sqrt 2-\sqrt 1|>ε$.
Therefore $g(x)=\sqrt x$ is uniformly continuous.

Answer 1

1. "The function $f$ is uniformly continuous if for each $ε>0$ there is some $δ>0$ such that $x,y∈A$ and $|x-y|<δ$ implies $|f(x)-f(y)|>ε$." - The last portion of this definition should actually say "$|x-y|<δ$ implies $|f(x)-f(y)|<ε$." You should also specify "uniformly continuous on a set A" since a function can be uniformly continuous on one set but not another.
2. "If we are looking for ε>0" - I believe that you mean "if we are given $\epsilon$". We are actually looking for $\delta$ given this $\epsilon.$
3. "... then we can substitute. $g(x)=\sqrt{2}$ and $g(y)=\sqrt{1}$." - This is where the most important error is. You cannot substitute particular values of $x$ and $y$ to get your proof. You have substituted $x = 2$ and $y =1$. The point of uniform continuity is that $|x-y|<δ$ implies $|f(x)-f(y)|<ε$ for any $x,y \in [1,2]$. That is, the choice of $\delta$ does not depend on $x$ and $y$.

Here is an example of a proof that a function is uniformly continuous.

Let $f$ be the function given by $f(x) = \frac{1}{x}$ for all $x \in [1,2]$. Suppose that $\epsilon > 0$ is given, and let $\delta = \epsilon.$ Then we have that for any $x,y \in [1,2]$, $$|x-y| < \delta \implies |f(x)-f(y)| = |\frac{1}{x}-\frac{1}{y}| = |\frac{y-x}{xy}| \leq |\frac{y-x}{1}| < \epsilon.$$ Since $\delta$ does not depend on $x$ and $y$, $f$ is uniformly continuous on $[1,2]$

Answer 2

Definition : $f$ is called uniformly continuous on $X$ if for all $ε > 0$ there exists a $δ > 0$ such that for all $x, y ∈ X, |x − y| < δ$ implies $|f(x) − f(y)| < ε$.

Claim:$f(x)= \sqrt x$ is uniformly continuous.

Proof:Let $\epsilon > 0.$ Pick $\delta = \epsilon^2.$ Then for $|x-y| < \delta$ we have

$$|\sqrt x - \sqrt y|^2 \leq |\sqrt x - \sqrt y||\sqrt x + \sqrt y| = |x-y| < \epsilon^2 \implies |\sqrt x - \sqrt y| < \epsilon. $$